What is the distribution of the ratio between independent Beta and Gamma random variables? What would be the distribution of the following equation:
$$y = \frac{a}{(a+d)^2}$$
where $a, d$ $\sim$ $\Gamma(M,c)$. Additionally, $a$ and $d$ are independent random variables.
 A: Francis has provided the definitive elegant solution. Following his beautiful method, it seems possible to produce a somewhat simpler form for the pdf.
In particular, following Francis' elegant solution, let $V \sim \text{Beta}(m,m)$ with pdf $f(v)$: 

.. and let $U \sim \text{Gamma}(2m, c)$ where $U$ is independent of $V$, and let $Z = \frac1U$, so that $Z \sim \text{InverseGamma}(2m,c)$ with pdf $g(z)$: 

Then the desired pdf of $Y = \frac{V}{U} = V Z$, say $h(y)$, can be found with:

where I am using the TransformProduct function from the mathStatica package for Mathematica to automate the nitty gritties.
So does it work?
Here is a quick Monte Carlo check of the:


*

*simulated pdf of $Y = \frac{A}{(A+D)^2}$  (generated from $A$ and $D$) (blue squiggly)

*compared to the exact theoretical pdf $h(y)$ (dashed red) derived above:

when parameter $m = 4$ and $c=3.01$. Looks fine.
A: For independent $a, d\sim\operatorname{\Gamma}(M,c)$, a remarkable result is that $U=a+d$ and $V=a/(a+d)$ are also independent. In addition, $U\sim\operatorname{\Gamma}(2M, c)$, $V\sim\mathrm{B}(M,M)$. See for example Ch.25, Sec. 2 of Johnson, Kotz, and Balakrishnan's 
Continuous Univariate Distributions, Volume 2 for some details.
Now this says $$Y=\frac{a}{(a+d)^2}=\frac{1}{a+d}\frac{a}{a+d}=\frac{V}{U}$$ is a ratio between independent Gamma and Beta random variables. Nadarajah and Kotz's On the Product and Ratio of Gamma and Beta Random Variables and Nadarajah's The Gamma Beta Ratio Distribution have discussed this distribution thoroughly.
EDIT: Using the CDFs and PDFs given in the articles mentioned above, we can derive (simplified with Wolfram Alpha) the CDF of $Y$ as: $$F_Y(y) = 1-\frac{\Gamma(3M)}{2(cy)^{2M}\Gamma(M+1)\Gamma(4M)}{}_2F_2\left(2M, 3M; 2M+1,4M; -\frac{1}{cy}\right),$$ and the PDF of $Y$ as: $$f_Y(y)=\frac{\Gamma(3M)}{y(cy)^{2M}\Gamma(M)\Gamma(4M)}{}_1F_1\left(3M; 4M; -\frac{1}{cy}\right)$$ where ${}_1F_1$ and ${}_2F_2$ are the generalized hyper-geometric functions. These expressions are in accordance with @wolfies's result.
