Answer
$\frac{∂f}{∂x}=e^{xy}+xye^{xy}$
$\frac{∂f}{∂y}=x^2e^{xy}$
$\frac{∂f}{∂x}|_{(1,-1)}=0$
$\frac{∂f}{∂y}|_{(1,-1)}=\frac{1}{e}$
Work Step by Step
We find the partial derivatives as follows:
$f(x,y)=xe^{xy}$
$\frac{∂f}{∂x}=e^{xy}+xe^{xy}(y)=e^{xy}+xye^{xy}$
$\frac{∂f}{∂y}=xe^{xy}(x)=x^2e^{xy}$
$\frac{∂f}{∂x}|_{(1,-1)}=e^{1(-1)}+1(-1)e^{1(-1)}=e^{-1}-e^{-1}=0$
$\frac{∂f}{∂y}|_{(1,-1)}=1^2e^{1(-1)}=e^{-1}=\frac{1}{e}$
