#### Answer

$$2\cos 2$$

#### Work Step by Step

Given $$g(x, y)=x^{2}-y^{2},\ \ \ \ \ \ x=e^{u} \cos v,\ \ \ y=e^u\sin v$$
Since at $(u,v)=(0,1)$; $(x,y)= (\cos(1),\sin(1))$
$$
\frac{\partial g}{\partial \:x}=2x,\ \ \ \ \ \frac{\partial g}{\partial \:y}=-2y \\ \frac{\partial x}{\partial u}=e^{u} \cos v,\ \ \ \ \ \ \ \ \ \ \ \frac{\partial y}{\partial u}=e^u\sin v
$$
Then
\begin{align*}
\frac{\partial g }{ \partial u}&=\frac{\partial g }{ \partial x}\frac{\partial x }{ \partial u}+\frac{\partial g }{ \partial y}\frac{\partial y }{ \partial u}\\
&= 2e^u (x\cos v -y \sin v)\\
&=2e^{2u}( \cos2v)
\end{align*}
Hence
\begin{align*}
\frac{\partial g }{ \partial u}\bigg|_{(0,1)}&= 2\cos 2
\end{align*}