$\text{My attempt:}$

$$\begin{align}x'&=\sin(x)\cos(y)=F(x,y)\\y'&=-\cos(x)\sin(y)=G(x,y)\end{align}$$

Let $$\begin{align}\cases{F(x,y)=\sin(x)\cos(y)=0\\G(x,y)=-\cos(x)\sin(y)=0}\end{align}$$

$$\implies\cases{x=k\pi,y=k\pi\\x={\pi\over2}+k\pi,y={\pi\over2}+k\pi}$$

where $k$ can only be $-1,0,1$, since we have $x\in(-4,4)$ and $y\in(-4,4)$, $\pi\cong 3.14159265359$. $\\$

Thus we have the following critical points:$\\$

Case#1: $(0,0);(\pi,0);(-\pi,0);(0,\pi);(0,-\pi);(\pi,\pi);(-\pi,\pi);(-\pi,-\pi);(\pi,-\pi)$. $\\$

Case#2: $({\pi\over2},{\pi\over2});(-{\pi\over2},-{\pi\over2});({\pi\over2},-{\pi\over2});(-{\pi\over2},{\pi\over2})$.

$$J=\begin{pmatrix}F_x(x,y)&F_y(x,y)\\G_x(x,y)&G_x(x,y)\end{pmatrix}=\begin{pmatrix}\cos(x)\cos(y)&-\sin(x)\sin(y)\\\sin(x)\sin(y)&-\cos(x)\cos(y)\end{pmatrix}$$

When $(x,y)=$ critical points in Case#1: we get diagonal matrices $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ or $\begin{pmatrix}-1&0\\0&0\end{pmatrix}$, wtih eigenvalues $\lambda=\pm1$, eigenvectors $\xi=\begin{pmatrix}1\\0\end{pmatrix}$ or $\begin{pmatrix}0\\1\end{pmatrix}$. $\\$

Thus, the critical points in Case#1 are all $\text{Saddle Points}$. $\\$

When $(x,y)=$ critical points in Case#2: we get matrices $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ or $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ with eigenvalues $\lambda=\pm i$, eigenvectors $\xi=\begin{pmatrix}1\\i\end{pmatrix}$ or $\begin{pmatrix}1\\-i\end{pmatrix}$. $\\$

Thus, the critical points in Case#2 are all $\text{Center}$. $\\$

Therefore based on the phase portraits for saddle point and center, we can conclude the phase portrait for the given system will be like that.