I have no idea how to calculate $z_x+z_y$ at a point $\left( \frac{\pi +3}{3}, \frac{\pi+1}{2}\right)$, if $z=uv^2$ and $x=u+sinv$, $y=v+cosu$. $z$ is not expressed in terms of $x$ and $y$. Maybe it is meant to be solved as $x_u=1$ and $x_v=cosv$, $y_u=-sinu$ and $y_v=1,$ then $x_u=cos^2v$ and $y_u=-sinu$.
$$z_u=v^2 \Rightarrow v^2= \frac{\partial z}{\partial x} \cdot 1+\frac{\partial z}{\partial y} \cdot (-sinu)$$
$$z_v=2uv \Rightarrow 2uv=\frac{\partial z}{\partial x} cosv+\frac{\partial z}{\partial y}\cdot 1$$
$$\Rightarrow \frac{\partial z}{\partial x}=v^2+sin(u) \frac{\partial z}{\partial y}$$
$$\Rightarrow 2uv=cos(v)\left( v^2+sin(u)\frac{\partial z}{\partial y} \right)+\frac{\partial z}{\partial y}$$
$$\Rightarrow 2uv-v^2cos(v)=\frac{\partial z}{\partial y}\left(sin(u)+1 \right)$$
$$\Rightarrow \frac{\partial z}{\partial y}=\frac{2uv-v^2cos(v)}{sin(u)+1}$$
I know that what I have done looks just confusing.
$$v^2=\frac{\partial z}{\partial x}+\frac{\sqrt{3}}{2}\Rightarrow \frac{\partial z}{\partial x}=\frac{\pi^2}{9}-\frac{\sqrt{3}}{2}$$
$$2uv=\frac{\partial z}{\partial y} \Rightarrow \frac{\partial z}{\partial y}=\frac{\pi^2}{18}$$
$$z_x+z_y=\frac{3\pi^2-9\sqrt{3}}{18}$$
Differentiate $u+\sin v=x$ and $v+\cos u=y$ with respect to $x$ and $y$, respectively,
$$u_x+\cos v\> v_x=1,\>\>\>\>\> v_x-\sin u\> u_x=0,\>\>\>\>\> u_y+\cos v\> v_y = 0,\>\>\>\>\> v_y-\sin u\> u_y = 1$$
Solve the linear equations,
$$u_x= \frac1{1+\sin u\cos v},\>\>\>\>\>v_x= \frac{\sin u}{1+\sin u\cos v}$$ $$u_y= -\frac{\cos v}{1+\sin u\cos v},\>\>\>\>\>v_y= \frac{1}{1+\sin u\cos v}$$
Then, from $z=uv^2$, evaluate
$$z_x+z_y =( z_u u_x+z_v v_x)+( z_u u_y+z_v v_y)=\frac{v^2(1-\cos v)+2uv(1+\sin u)}{1+\sin u\cos v}\tag 1$$
Given the point $(x,y)=(\frac{\pi+3}3,\frac{\pi+1}2)$, set
$$\frac{\pi+3}3=u+\sin v,\>\>\>\>\>\frac{\pi+1}2 = v+\cos u$$
to get $(u,v) = (\frac\pi3,\frac\pi2)$. Plug the point into the expression (1) to obtain the result
$$z_x+z_y =\frac{7+2\sqrt3}{12}\pi^2$$