Answer
$$f_x= y\cos(xy)e^{-x-y}-\sin(xy)e^{-x-y},$$
$$f_y= x\cos(xy)e^{-x-y}-\sin(xy)e^{-x-y}.$$
Work Step by Step
To calculate the partial derivatives, treat all variables as constant except the variable we are deriving with respect to.
Since $f(x,y)=\sin(xy)e^{-x-y}$, then we have
$$f_x=\frac{\partial f}{\partial x}=y\cos(xy)e^{-x-y}-\sin(xy)e^{-x-y},$$
$$f_y=\frac{\partial f}{\partial y}=x\cos(xy)e^{-x-y}-\sin(xy)e^{-x-y}.$$