Answer
$g_{xx}=2y-y \sin \space x ; \\g_{yy}=-\cos y$
and $g_{xy}=g_{yx}=2x+\cos \space x$
Work Step by Step
Our aim is to take the first partial derivative of the given function $f(x,y)$ with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa:
$g_x=2x \space y+y \cos x ;g_y=x^2-\sin y+\sin x$
Next, take the derivative of the first order partial derivatives to compute the second order partial derivatives.
$g_{xx}=2y-y \sin \space x ; \\g_{yy}=-\cos y$
Now, $g_{xy}=g_{yx}=2x+\cos \space x$
