nonlinear regression two equivalent models on paper, but different estimated parameters I measured one response variable
Y1

as a function of two measured independent variables
X1 and X2

It is common practice in my field of research to transform Y1 into another response variable: 
Y2 = 10*X1 * ((-1)+1 / Y1)

Then to fit the following model (M1):
Y2 ~ a * X1^b * X2^c + error

in order to estimate (a, b, c).
However, it is also possible to estimate these three parameters witout transforming Y1 into Y2, using this equivalent model (M2) (equivalent on paper at least):
Y1 ~ 10*X1 / (a * X1^b * X2^c + 10*X1) + error

The problem is, these two ways of estimating (a, b, c) yields different estimates. I personnally believe M2 is to be preffered over M1, because calculating Y2 from Y1 and X1 introduces uncertainties from X1 into the nonlinear response variable.
I wish / need to know which model (M1 or M2) to use, and why.
Thanks in advance.
PS: I did not present any reproducible example nor graphs because I think this question is sufficiently general and clear, and does not need it. Thanks for your understanding.
 A: There are NOT equivalent least squares models. The error in one model is a transformation of the error in the other. Whichever model's error is closer to being Normally distributed should be the better model.  Edit: see whuber's answer for more elaboration on the error transformation.
There is a second matter as well, and that is the question of what numerical solution is obtained by the nonlinear least solution algorithm. The solution obtained can depends on the algorithm used to solve it as well as the starting value (initial guess) for the parameters being estimated. Depending on the algorithm and starting value, it's possible that the algorithm will exit without finding a local optimum. It's possible that a local optimum will be found which is not the global optimum. 
You should want the globally optimal solution.  Whether you find it is another matter.  That's where it helps to know what you're doing in nonlinear optimization, which unfortunately most people performing nonlinear least squares don't.
A: Model 2 is
$$Y_1 = \frac{10 X_1}{a X_1^b X_2^c + 10X_1} + \delta$$
whereas Model 1 is
$$10X_1 \left(-1 + \frac{1}{Y_1}\right) = a X_1^b X_2^c + \varepsilon,$$
which can be solved for $Y_1$ to read
$$Y_1 = \frac{10 X_1}{a X_1^b X_2^c + 10X_1 + \varepsilon}.$$
Implicitly it is supposed the errors $\varepsilon$ or $\delta$, as the case may be, are independent, have identical distributions, and are centered at zero.  
To compare the two models let's assume the variability of $\varepsilon$ is substantially less than the magnitude of $aX_1^bX_2^c + 10X_1$.  We may then use the Binomial Theorem (or, equivalently, a Taylor series) to approximate the right hand side of Model 1 (to first order in $\varepsilon$) as
$$Y_1 \approx \frac{10 X_1}{a X_1^b X_2^c + 10X_1}\left(1 - \frac{\varepsilon}{a X_1^b X_2^c + 10X_1} + \cdots\right).$$
Comparing to Model 2, we see the difference between them lies in the error terms:
$$\delta \approx \frac{-10 X_1}{\left(a X_1^b X_2^c + 10X_1\right)^2} \varepsilon.$$
These are different models because if the $\varepsilon$ have identical distributions, the $\delta$ cannot--since they rescale the $\varepsilon$ by factors that depend on the variables $X_1$ and $X_2$. Conversely, if the $\delta$ have identical distributions then the $\varepsilon$ cannot.
To decide which one to use (if either), you will need additional information concerning the distributions of the errors.  This can be obtained in many ways, including


*

*Theoretical considerations.  For instance, if the error is intended to represent measurement variability of $Y_1$ and that variability is known to be (roughly) constant across a range of values of $Y_1$, the Model 2 is a good choice.

*Analysis of repeated measurements.

*Review of diagnostic information from each model (related to the possible heteroscedasticity of the residuals).


The red curves show the correct underlying relationships.  The dots show simulated data.  Their vertical deviations from the red curves represent the errors. The dispersion of the errors in Model 1, at the left, visibly varies with the independent variables. The dispersion in Model 2, at the right, does not.
This figure shows data simulated with the R code below.  To simplify the presentation, all values of $X_2$ were set to a constant value, causing all variation in $Y_1$ to be associated with variation in $X_1$ only.  This simplification does not change the nature of the differences between the two models.
a <- 1
b <- 2
c <- 3
n <- 250
sigma <- 2
#
# Generate data according to two models.
#
set.seed(17)
x1 <- rgamma(n, 2) + 1
x2 <- rep(1, n)
epsilon <- rnorm(n, sd=sigma)
y.m1 <- 10 * x1 / (a * x1^b * x2^c + 10*x1 + epsilon)

# (Make them have comparable errors on average.)
tau <- mean(abs(-10 * x1 / (a * x1^b * x2^c + 10*x1)^2))
delta <- rnorm(n, sd=tau)
y.m2 <- 10 * x1 / (a * x1^b * x2^c + 10*x1) + delta
#
# Plot the simulated data.
#
reference <- function() curve(10 * x / (a*x^b + 10*x), add=TRUE, col="Red", lwd=2)
par(mfrow=c(1,2))
plot(x1, y.m1, main="Model 1", xlab="X1", ylab="Y1", col="#00000070")
reference()
plot(x1, y.m2, main="Model 2", xlab="X1", ylab="Y1", col="#00000070")
reference()

