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The first Google hit says:

enter image description here

This highly cited answer says:

Partial r is just another way of standardizing the coefficient, along with beta coefficient (standardized regression coefficient)$^1$. So, if the dependent variable is $y$ and the independents are $x_1$ and $x_2$ then

$$\text{Beta:} \quad \beta_{x_1} = \frac{r_{yx_1} - r_{yx_2}r_{x_1x_2} }{1-r_{x_1x_2}^2}$$

$$\text{Partial r:} \quad r_{yx_1.x_2} = \frac{r_{yx_1} - r_{yx_2}r_{x_1x_2} }{\sqrt{ (1-r_{yx_2}^2)(1-r_{x_1x_2}^2) }}$$

Those are not compatible definitions. Does anyone have a source for the formula in the CrossValidated answer?

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    $\begingroup$ These are defining different things. The two quotations just happen to use the word "standardizing" in different senses, but both are perfectly clear. $\endgroup$
    – whuber
    Commented Apr 22 at 13:47

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They are compatible in that you can show that the "standardized coefficient" $\hat{\beta^*}$, defined by your first quotation, is indeed identical to the expression $\hat{\beta}_{x_1}$ in your cited answer. This has been proven in my answer as an intermediate step for validating the equivalence between two definitions of $R^2$. Although your second quotation involved the concept of "partial R", I am assuming it is actually irrelevant because your first quotation solely concerns with the "standardized regression coefficient".

Specifically, using the notations in my answer, I showed that for a general multiple linear regression with $p$ independent variables (where $\hat{\gamma}$ denotes the OLS estimate for the original regression parameters): \begin{align*} & \hat{\beta^*} = s_Y^{-1}\Lambda\hat{\gamma}, \tag{1}\label{1} \\ & \hat{\beta^*} = r_{XX}^{-1}r_{YX}, \tag{2}\label{2} \end{align*} where $\Lambda = \operatorname{diag}(s_{X_1}, \ldots, s_{X_p})$, $r_{XX}$ is the order $p$ sample correlation matrix of $(X_1, \ldots, X_p)$, $r_{YX}$ is the $p$-long column vector of sample correlations between $Y$ and $(X_1, \ldots, X_p)$, and $s_{X_i}, s_Y$ are sample standard deviations of $X_i$ and $Y$. In your case, $p = 2$, and what you are trying to confirm is just that the first component of $\eqref{1}$ and the first component of $\eqref{2}$ are the same (which is of course true given that as two vectors, $\eqref{1}$ and $\eqref{2}$ are proven to be the same).

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