Using the graph it's very clear that the points are $(0,0)$ and $(\pi/2,0)$, but I don't know how to solve it algebraically.
What I have done: I take $y=0$ in the given equation, and find the corresponding $x$ values, but I don't know if it's the right way, because using this I can't verify whether it also satisfies for other $y$ values.
Please help!
For every $t\in[0,\frac{\pi}{2}]$ : $\sin(t)-\sin^2(t)\ge0$.
Hence, for every $(x,y)\in[0,\frac{\pi}{2}]^2$ :
$$\sin(x)-\sin^2(x)+\sin(y)-\sin^2(y)=0\implies\left(\sin(x)=\sin^2(x)\quad\mathrm{and}\quad\sin(y)=\sin^2(y)\right)$$
Finally, solutions are $(0,0)$, $(\frac{\pi}{2},0)$, $(0,\frac{\pi}{2})$ and $(\frac{\pi}{2},\frac{\pi}{2})$.
Hint: $\sin^2 x \le \sin x, \sin^2 y \le \sin y \implies \sin^2 x = \sin x, \sin^2 y = \sin y$.