Using the "transformation method", I am to show that with $W,W_1,W_2 \stackrel{i.i.d}{=} N(0,1)$, a random variable $Z \sim BVN(0,0;1,1;\rho)$ is identically distributed to the random variable given by
$$(Z_1,Z_2)\stackrel{d}{=}(W\sqrt{\rho}+W_1\sqrt{1-\rho}, W\sqrt{\rho}+W_2\sqrt{1-\rho}).$$
Generalizing what I am given as the "Bivariate transformation method" onto $n$ dimensions, I take it to be given as: Let $Z$ be a $n$-dimensional random variable and let $h_i$ be bijective functions mapping random variables $U_i$ with joint distribution function $f_U(u_1,\dots,u_n)$ onto the $Z_i$. If the inverse functions $h_i$ have continuous partial derivatives in regard to the $Z_i$, and the determinant of the Jacobian of the inverse is $\neq 0$, then the joint density function of $Z$ is given by:
$$f_Z(z_1,\dots,z_n)=f_U(h_1^{-1}(z_1,\dots,z_n),\dots,h_n^{-1}(z_1,\dots,z_n))\lvert J \rvert$$
Because the $h_i$ are functions in three variables, I add one variable to the multivariate $Z$, so that my Jacobian will be quadratic. I let $Z_3 := W\sqrt{\rho} $ and obtain
$$h_1^{-1}=\frac{Z_1-Z_3}{\sqrt{1-\rho}}\\ h_2^{-1}=\frac{Z_2-Z_3}{\sqrt{1-\rho}}\\ h_3^{-1}=\frac{Z_3}{\sqrt{\rho}}$$
I then take the joint distribution $f_W$ of three $N(0,1)$ RVs, insert the inverse functions of the $h_i$ and attempt to obtain the target density (a $BNV(0,0;1,1;\rho)$ RV). But I fail to get there. Is my approach correct and I am likely miscalculating somewhere, or is something about my approach wrong?
Further note: My aim is to (once I have the joint density $f_Z$) integrate over the $z_3$, obtaining what should be the density for a $BVN(0,0;1,1;\rho)$ random variable. I end up with something that looks very hard to integrate though, and so I assume that somewhere in the process I have been going wrong.
Recall that an affine transform of a normal random variable is also normal: $$X\sim N(\mu,\Sigma)\implies a+bX\sim N(a+b\mu,b\Sigma b').$$
Now note $(W,W_1,W_2)'\sim N(0,I_3)$ and write
$$\left(\begin{array}{c} W\sqrt{\rho}+W_{1}\sqrt{1-\rho}\\ W\sqrt{\rho}+W_{2}\sqrt{1-\rho} \end{array}\right)=\left(\begin{array}{ccc} \sqrt{\rho} & \sqrt{1-\rho} & 0\\ \sqrt{\rho} & 0 & \sqrt{1-\rho} \end{array}\right)\left(\begin{array}{c} W\\ W_{1}\\ W_{2} \end{array}\right).$$