<s>[INST] Let $z=\ln \tan\frac xy.$ What is $z_x$ and what is $z_y$? Let $$z=\ln \tan\frac xy.$$ What is $z_x$ and what is $z_y$?
Thanks ahead:)
What I have tried:
$$z_x=\frac{1}{\tan \frac xy} \frac{1}{1+(\frac xy)^2} \frac 1y=\frac {y}{\tan \frac xy (x^2+y^2)}$$
$$z_y=\frac {-x}{\tan \frac xy (x^2+y^2)}.$$
I am not sure I am right. Help, help, help...
 [/INST] As pointed out in comments, you misread $\tan$ as $\tan^{-1}$.
The answers are:
$$
\begin{array}{c}
z_x = \frac{1}{\tan\frac{x}{y}}\frac{1}{y}\sec^2\frac{x}{y}=\frac{\sec\frac{x}{y}\csc\frac{x}{y}}{y} \\
z_y = -\frac{1}{\tan\frac{x}{y}}\frac{x}{y^2}\sec^2\frac{x}{y}=\frac{-x \sec\frac{x}{y}\csc\frac{x}{y}}{y^2}
\end{array}
$$
</s>