Below is the question:
$$\begin{cases} v+log\left|u\right|=xy \\ u+log\left|v\right|=x-y \end{cases}\implies \begin{cases} \frac{1}{u}\frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{x}}=y \\ \frac{1}{v}\frac{\partial{v}}{\partial{x}}+\frac{\partial{u}}{\partial{x}}=1 \end{cases}\implies \frac{\partial{u}}{\partial{x}}=\frac{\begin{vmatrix} yu & u \\ v & 1 \\ \end{vmatrix}}{\begin{vmatrix} 1 & u \\ v & 1 \\ \end{vmatrix}}=\frac{u\left(y-v\right)}{1-uv}$$
and this is the equivalent of the first partial derivative. Now I have to find $\frac{\partial^2{u}}{\partial{x}}$. I think it looks like:
$$\frac{\partial^2{u}}{\partial{x}}=\frac{\partial}{\partial{x}}\left(\frac{u\left(y-v\right)}{1-uv}\right)=\frac{\left(1-uv\right)\left(u\left(-1\right)+\left(y-u\right)\right)-u\left(y-u\right)\left(-v\right)}{\left(1-uv\right)^2}$$ $$=\boxed{\frac{\left(1-uv\right)\left(y-2u\right)+\left(yuv-vu^2\right)}{\left(1-uv^2\right)}}$$
Am I right? Or do I have to do a determinant like they did (even if I am right, how would you solve this using a determinant?).
Thank you,
Brandon