The Square of

$\begin{aligned} y = x^2 & = (x+y)^2 \\ \implies (x+y)^2 & = x^2 \\ x^2 + 2xy + y^2 & = x^2 \\ 2x\color{#3D99F6}{y} + \color{#3D99F6}{y}^2 & = 0 \quad \quad \small \color{#3D99F6}{y = x^2} \\ 2x(\color{#3D99F6}{x^2}) + (\color{#3D99F6}{x^2})^2 & = 0 \\ 2x^3 + x^4 & = 0 \\ x^3 (2+x) & = 0 \\ \implies x & = \begin{cases} 0 & \implies y = 0^2 = 0 \\ -2 & \implies y = (-2)^2 = 4 \end{cases} \end{aligned}$

The sum of values of $y$ $= 0 + 4 = \boxed{4}$

First off, let's start with the last 2 parts of the given equality, $x^2$ and $(x+y)^2$ . Expanding $(x+y)^2$ , and simplifying, we have $0=y(2x+y)$ . Therefore $y=0$ . Then we take the other half, $2x+y=0$ , and compare it to $y=x^2$ . Adding the equations to each other and simplifying, we have $x(x+2)=0$ . If $x=0$ , then we have $y=0$ . If $x=-2$ , then $y=4$ . If we plug these values back into the equation, we can see that all values work, and so the sum of our two answers is equal to $4$ .