Let $P(x, y)$ be the given assertion,
$$P(0,1): f(0) = 0$$
Assume that there exists $a \neq 0$ such that $f(a) = 0$. Then $P(x, a)$:
$$xf(x) + af(xa) = xf(x) \Rightarrow \forall x \in \mathbb{R} \ f(xa) = 0 \Rightarrow \forall x \in \mathbb{R} \ f(x) = 0$$
And we found the first solution.

Let us now assume that $f$ is not identically zero, so $f(x) = 0 \Leftrightarrow x = 0$
$$P(-1,1): 0 = -f(-1) + f(-1) = -f(-1 + f(1)) \Rightarrow f(1) = 1$$
$$P(x, 1): xf(x) + f(x) = xf(x+1) \Rightarrow (x+1)f(x) = xf(x+1) \Rightarrow \forall x \neq 0, -1$$
$$\frac{f(x)}{x} = \frac{f(x+1)}{x+1} \quad (1)$$
We already have that $f(1) = 1$, so we can get from (1) that $\forall n \in \mathbb{N} \ f(n) = n$. Also, we can obtain from (1) that $f(-2) = 2f(-1)$.
$$P(-1,2): -f(-1) + 2f(-2) = -f(-1 + 2f(2)) = -f(3) = -3 \Rightarrow f(-1) = -1$$
By combining this with (1) we can get that $\forall n \in \mathbb{Z} \ f(n) = n$.

Let us now show that $\forall n \in \mathbb{Z}$ and $\forall x \in \mathbb{R}$, $(x+n)f(x) = xf(x+n)$. If $x \in \mathbb{Z}$ this is obvious, otherwise we can write that $\frac{f(x)}{x} = \frac{f(x+1)}{x+1} = \cdots = \frac{f(x+n)}{x+n}$ and obtain this equality.

$$\begin{align*} P(x, n): \quad xf(x) + nf(nx) &= xf(x + nf(n)) \\ &\Rightarrow xf(x) + nf(nx) = xf(x + n^2) = (x + n^2)f(x) \\ &\Rightarrow xf(x) + nf(nx) = xf(x) + n^2f(x) \\ &\Rightarrow f(nx) = nf(x). \tag{2} \end{align*}$$

After we'll divide $P(x, y)$ by $x \neq 0$:
$$f(x) + \frac{yf(xy)}{x} = f(x + yf(y)). \quad (4)$$

$P(1, y): 1 + yf(y) = f(1 + yf(y))$

Let us now put $x = yf(y)$ in (1), so $\frac{f(yf(y))}{yf(y)} = \frac{f(yf(y)+1)}{yf(y)+1} = 1 \Rightarrow$
$$f(yf(y)) = yf(y). \quad (3)$$

Put $x = xf(x)$ there and by symmetry we'll get the following:
$$\begin{align*} f(xf(x)) + \frac{yf(xyf(x))}{xf(x)} &= f(xf(x) + yf(y)) \\ &= f(yf(y)) + \frac{xf(xyf(y))}{yf(y)} \\ \Rightarrow f(xf(x)) - \frac{xf(xyf(y))}{yf(y)} &= f(yf(y)) - \frac{yf(xyf(x))}{xf(x)} \end{align*}$$
If we put here $2x$ instead of $x$ there, then the LHS will increase by 4 times because of (2) and the RHS will not change, so both are zero and hence, $f(xf(x)) = \frac{xf(xyf(y))}{yf(y)} \Rightarrow$
$$f(xyf(y)) = f(x)yf(y) \quad (5)$$

Let us put $x = x + 1$ into (5):
$$\begin{align*} f(x+1)yf(y) &= f((x+1)yf(y)) = f(xyf(y) + yf(y)) \\ &\stackrel{(4)}{=} f(xyf(y)) + \frac{yf(xy^2f(y))}{xyf(y)} \\ &\stackrel{(5)}{=} f(x)yf(y) + \frac{y^2f(y)f(xy)}{xyf(y)} = f(x)yf(y) + \frac{yf(xy)}{x} \end{align*}$$
So, $(f(x+1) - f(x))x = \frac{f(xy)}{f(y)}$ and by substituting $y=1$ here, we are getting that
$$(f(x+1) - f(x))x = f(x) \Rightarrow f(xy) = f(x)f(y) \quad (6)$$
By using (6) in (3) we get that $f(f(y)) = y$. So
$$\begin{align*} P(f(y), y): \quad f(y)y + yf(f(y)y) &= f(y)f(f(y) + yf(y)) \\ &\Rightarrow f(y)y + y^2f(y) = yf(y)f(y+1) \\ &\Rightarrow f(y+1) = y+1 \Rightarrow \forall x \in \mathbb{R} \ f(x) = x \text{ is the second solution.} \end{align*}$$