Answer
$f_{xx}(x, y)=2m^{2}\cos(2mx+2ny)$,
$f_{xy}(x, y)=2mn \cos(2mx+2ny)$,
$f_{yx}(x, y)=2mn \cos(2mx+2ny)$,
$f_{yy}(x, y)=2n^{2}\cos(2mx+2ny)$.
Work Step by Step
$f(x, y)=\sin^{2}(mx+ny)$
$f_{x}(x, y)=\displaystyle \frac{\partial}{\partial x}[f(x, y)] \stackrel{\text{chain rule} }{=} 2\sin(mx+ny) \cos(mx+ny)\cdot m$
$...$ use: $\sin 2\alpha=2\sin\alpha\cos\alpha$
$f_{x}(x, y)=m\sin(2mx+2ny)$
$f_{xx}(x, y)=\displaystyle \frac{\partial}{\partial x}[f_{x}(x, y)] \stackrel{\text{chain rule} }{=} m\cos(2mx+2ny) (2m)$
$=2m^{2}\cos(2mx+2ny)$
$f_{xy}(x, y)=\displaystyle \frac{\partial}{\partial y}[f_{x}(x, y)] \stackrel{\text{chain rule} }{=} m\cos(2mx+2ny) (2n)$
$=2mn \cos(2mx+2ny)$
$f_{y}(x, y)=\displaystyle \frac{\partial}{\partial y}[f(x, y)] \stackrel{\text{chain rule} }{=} 2\sin(mx+ny) \cos(mx+ny)\cdot n$
$...$ use: $\sin 2\alpha=2\sin\alpha\cos\alpha$
$f_{y}(x, y)=n \sin(2mx+2ny)$
$f_{yx}(x, y) =\displaystyle \frac{\partial}{\partial x}[f_{y}(x, y)] \stackrel{\text{chain rule} }{=} n \cos(2mx+2ny) (2m)$
$=2mn \cos(2mx+2ny)$
$f_{yy}(x, y)=\displaystyle \frac{\partial}{\partial y}[f_{y}(x, y)] \stackrel{\text{chain rule} }{=} n \cos(2mx+2ny) (2n)$
$=2n^{2}\cos(2mx+2ny)$
