Partial Derivative of arctan Given that $$f(x,y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$
Find $f_x(x,y)$
My attempt,
$$
\begin{aligned}
f_x(x,y)&=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\cdot\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\\
&=\frac{1+y^2}{(1-xy)^2+(x+y)^2}\\
&=\frac{1+y^2}{1+y^2+x^2+x^2y^2}
\end{aligned}
$$
But the given answer is $\frac{1}{x^2+1}$.
How?

Because
$$\frac{1+y^2}{1+y^2+x^2+x^2y^2}$$
and
$$\frac{1}{x^2+1}$$
are the same thing.
