Answer
$z=-4y-4$
Work Step by Step
We find the first derivatives of the given function:
$ f_x (x,y)=y^2 \cos x$
$ f_y (x,y) =2y \sin x$
So $f_{x}\left( \dfrac{\pi }{2},-2\right) =0$ and $f_{y}\left( \dfrac{\pi }{2},-2\right) =-4$
The equation of the tangent line at $\left( \dfrac{\pi }{2},-2,4\right) $ is
$$ z-4 = f_{x}\left( \dfrac{\pi }{2},-2\right) \cdot \left( x-\dfrac{\pi }{2}\right)
$$ $$\Rightarrow z-4=0\left( x-\dfrac{\pi }{2}\right) +\left( -4\right) \left( y-\left( -2\right) \right)\\
\Rightarrow z=-4y-4$$
