Answer
$u_{xy}=e^{xy}\sin y(xy+1)+ye^{xy}\cos y=u_{yx}$
Work Step by Step
Consider the function $u=e^{xy}\sin y$
Need to prove the conclusion of Clairaut’s Theorem holds, that is, $u_{xy}=u_{yx}$
In order to find this differentiate the function with respect to $x$ keeping $y$ constant.
$u_{x}=ye^{xy}\sin y$
Differentiate the function with respect to $y$ keeping $x$ constant.
$u_{y}=xe^{xy}\sin y+e^{xy}\cos y$
Differentiate $u_{x}$ with respect to $y$ keeping $x$ constant.
$u_{xy}=\frac{∂}{∂y}[ye^{xy}\sin y]$ (apply product rule)
$=e^{xy}\sin y\frac{∂y}{∂y}+y\sin y\frac{∂e^{xy}}{∂y}+ye^{xy}\frac{∂(\sin y)}{∂y}$
$=e^{xy}\sin y+y\sin y(xe^{xy})+ye^{xy}\cos y$
Thus, $u_{xy}=e^{xy}\sin y(xy+1)+ye^{xy}\cos y$
Differentiate $u_{y}$ with respect to $x$ keeping $y$ constant.
$u_{yx}=\frac{∂}{∂x}[xe^{xy}\sin y+ye^{xy}\cos y]$
$=\frac{∂}{∂x}[xe^{xy}\sin y]+\frac{∂}{∂x}[ye^{xy}\cos y]$
$=e^{xy}\sin y\frac{∂}{∂x}[x]+x\times\frac{∂}{∂x}[e^{xy}\sin y]+ye^{xy}\cos y$
$=e^{xy}\sin y+xy\sin y(e^{xy})+ye^{xy}\cos y$
Thus, $u_{yx}=e^{xy}\sin y(xy+1)+ye^{xy}\cos y$
Therefore, $u_{xy}=e^{xy}\sin y(xy+1)+ye^{xy}\cos y$ and $u_{yx}=e^{xy}\sin y(xy+1)+ye^{xy}\cos y$
Hence, $u_{xy}=u_{yx}$
