Solution:
First we exclude all pairs $(x, y)$ that make the denominator vanish, that is, we impose the conditions $x \neq 0$, $y \neq 0$ and $x+y \neq 0$.
Under these conditions the expression $x y(x+y)$ is different from $0$, so, multiplying both sides of the given relation by it, we obtain that it is equivalent to $4 x y = (x+y)^2$, that is, to $(x-y)^2 = 0$, that is, to $x = y$.
Therefore the pairs sought are all and only those satisfying $x = y$, from which however the pair $(0,0)$ must be removed.

SECOND SOLUTION
The given relation is homogeneous, that is, it remains unchanged if $(\lambda x, \lambda y)$ is put in place of $(x, y)$, with $\lambda \neq 0$.
This means that it holds at a point $\left(x_0, y_0\right)$ different from the origin if and only if it holds at all points of the line passing through the origin and through $\left(x_0, y_0\right)$ (except the origin).

Substituting $m x$ in place of $y$ in the given relation, we obtain
$$\frac{4}{x+m x}=\frac{1}{x}+\frac{1}{m x}$$
which, since $x \neq 0$, becomes
$$\frac{4}{1+m}=1+\frac{1}{m}$$
which has as its only solution $m=1$.
This means that the set of all and only the solutions of the given relation consists of the points of the line $y=x$ minus $(0,0)$.

THIRD SOLUTION
After recognizing the homogeneity of the relation by proceeding as in the second solution, to show that the only good line is $y=x$, one can also check that by intersecting the given relation with another one, such as $x^2+y^2=1$, which intersects all the lines through the origin, one obtains only solutions with equal abscissa and ordinate.