Solution:
Plugging in $x = y \neq 0$ yields $f(f(x)) = \alpha x$ for some constant $\alpha$. This holds for all $x \neq 0, -1$. If $\alpha$ was any real number different from $0$ or $1$, we would get $f(f(-1/\alpha)) = -1$, which is a contradiction. Hence, only $\alpha = 0$ or $\alpha = 1$ are possible. In the former case $x = 1, y = 0$ gives a contradiction, so $f(f(x)) = x$ for all $x \in D, x \neq 0$.

Letting $x = 1, y = f(1)$ gives $f(1) = 0$ and so $f(0) = 1$. Now, for all other $x$, we can plug in $\frac{1}{x}$ instead of $x$ and $y = 1$ to obtain $f(x) = \frac{1-x}{1+x}$ (this formula now also works for $x = 0$), and we can verify that this is indeed a solution:
$$f(f(x)) = \frac{1 - \frac{1-x}{1+x}}{1 + \frac{1-x}{1+x}} = \frac{1-x - (1-x)}{1+x + 1-x} = x$$
and thus
$$(f(f(x)) + y) f\left(\frac{y}{x}\right) + f(f(y)) = (x + y) f\left(\frac{y}{x}\right) + y = (x + y) \frac{1 - \frac{y}{x}}{1 + \frac{y}{x}} + y = x - y + y = x$$