Answer
$g_u(0,0)=7,g_v(0,0)=2$
Work Step by Step
Notice that $g(0,0)=f(e^0+\sin0,e^0+\cos0)=f(1,2)$.
We will use the Chain Rule to find $g_u(0,0)$ and $g_v(0,0)$:
$$g_u(0,0)=f_x(1,2)\left.\frac{\partial}{\partial u}(e^u+\sin v)\right|_{(u,v)=(0,0)}+f_y(1,2)\left.\frac{\partial}{\partial u}(e^u+\cos v)\right|_{(u,v)=(0,0)}=
2\cdot e^0+5\cdot e^0=7$$
$$g_v(0,0)=f_x(1,2)\left.\frac{\partial}{\partial v}(e^u+\sin v)\right|_{(u,v)=(0,0)}+f_y(1,2)\left.\frac{\partial}{\partial v}(e^u+\cos v)\right|_{(u,v)=(0,0)}=
2\cdot\cos0+5\cdot(-\sin 0)=2$$
