Suppose there exists $2$ fixed points, $x$ and $y$. Then $x + y > 0$, so we get
$$
x^2(x + y) = (x + y)f(xy)
$$
and therefore $f(xy) = x^2$. Swapping $x$ and $y$, we get $f(xy) = y^2$. Therefore $x^2 = y^2$, and $x = y$.
So, there can be at most $1$ fixed point. Now, you have shown that $1$ is a fixed point. But $xf(x)$ is a fixed point for all $x > 0$, therefore
$$
f(x) = \frac{1}{x}
$$
for all $x \in \mathbb R^+$. We must then check this solution. For all $x,y$ we have
$$
\begin{aligned}
x^2(f(x) + f(y)) &= x^2\left(\frac 1 x + \frac 1 y\right)\\
&= x + \frac{x^2}{y}\\
&= \frac{x}{y}\left(y + x\right)\\
&= (x + y)f\left(\frac{y}{x}\right)\\
&= (x + y)f(f(x)y)
\end{aligned}
$$
So $f\colon x \mapsto \frac{1}{x}$ is a solution, and is the only solution.
It is usually pretty hard to prove monotonicity in functional equations, without outright finding the function. To reduce the possible number of fixed points, it is however pretty common to plug in two fixed points into the equation, which makes the contibutions of $f$ "vanish".
You can show that the functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \big( x + f ( y ) \big) = y + f ( x + 1 ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ are exactly those of the form $ f ( x ) = A ( x ) + 1 $, where $ A $ is an additive involution, i.e. $ A ( x + y ) = A ( x ) + A ( y ) $ and $ A \big( A ( x ) \big) = x $ for all $ x , y \in \mathbb R $. If further regularity conditions like continuity, local boundedness, integrability or measurability are assumed for $ f $, then $ A $ will be regular, too. The only regular additive functions are those of the form $ A ( x ) = c x $ for some constant $ c \in \mathbb R $ (see here). Thus the only regular additive involutions are $ A ( x ) = x $ and $ A ( x ) = - x $, as we must have $ A \big( A ( 1 ) \big) = c ^ 2 = 1 $. Therefore, the only regular solutions to \eqref{0} are $ f ( x ) = x + 1 $ and $ f ( x ) = - x + 1 $. Without any further conditions on $ f $, one can use the axiom of choice to show that there are nonregular solutions, too (see Example 5.6 in this PDF file).
It's straightforward to check that if $ f $ is of the form $ f ( x ) = A ( x ) + 1 $ for some additive involution $ A $, then it satisfies \eqref{0}. We try to prove the converse.
Letting $ x = 0 $ in \eqref{0} we get
$$ f \big( f ( y ) \big) = y + f ( 1 ) \text . \tag 1 \label 1 $$
In particular, \eqref{1} shows that if $ f ( x ) = f ( y ) $ then $ x = y $. Letting $ y = 0 $ in \eqref{1} shows that $ f \big( f ( 0 ) \big) = f ( 1 ) $, and thus by injectivity, $ f ( 0 ) = 1 $. Hence, putting $ x = - 1 $ in \eqref{0} we have
$$ f \big( f ( y ) - 1 \big) = y + 1 \text . \tag 2 \label 2 $$
\eqref{2} shows that if we substitute $ x - 1 $ for $ x $ and $ f ( y ) - 1 $ for $ y $ in \eqref{0}, we get
$$ f ( x + y ) = f ( x ) + f ( y ) - 1 \text . \tag 3 \label 3 $$
Now, if we define $ A : \mathbb R \to \mathbb R $ with $ A ( x ) = f ( x ) - 1 $, then by \eqref{3} we can see that $ A $ is additive, and by \eqref{2} we can see that $ A $ is an involution. This proves what was desired.
Best Answer
Suppose $f:\Bbb R\to \Bbb R$ satisfies $$f\big(f(x)-x+y^2\big)=yf(y)\tag{1}$$ for all $x,y\in \Bbb R$. Plugging in $x,y=1$ in $(1)$, we get $$f\big(f(1)\big)=f(1).$$ Hence, substituting $x=f(1)$ in $(1)$, we obtain $$f(y^2)=yf(y).\tag{2}$$ This shows that $f(0)=0$ and for $y\ne 0$, $$-yf(-y)=f\big((-y)^2\big)=f(y^2)=yf(y),$$ so $f(-y)=-f(y)$. Hence $f$ is an odd function.
From $(1)$ and $(2)$, we get $$f\big(f(x)-x+y^2\big)=f(y^2)$$ for all $x,y\in\mathbb{R}$. Substitute $-x$ for $x$ in the last equation and using the fact that $f$ is odd, we get $$f\big(f(x)-x-y^2\big)=f(-y^2)$$ for all $x,y\in\mathbb{R}$. Hence, $$f\big(f(x)-x+y)=f(y)\tag{3}$$ for all $x,y\in\mathbb{R}$.
Define $P$ to be the additive subgroup of $\mathbb{R}$ generated by $\big\{f(x)-x\big|x\in\Bbb R\big\}$. Then, we see that for any $z\in\mathbb{R}$ and $p\in P$, we have $$f(z+p)=f(z).\tag{4}$$ Consequently $f$ is an odd periodic function which is invariant under translation by elements of $P$.
We claim that $P=f^{-1}(0)$. First if $p\in P$, then $f(p)=f(p+0)=f(0)=0$, so $p\in f^{-1}(0)$. Conversely, suppose that $p\in f^{-1}(0)$, then with $x=p$ in $(3)$, we have $$f(-p+y)=f\big(f(p)-p+y\big)=f(y)$$ so that $p\in P$.
For each $p\in P$, we see that $$f\left(\left(\frac{1+p}{2}\right)^2\right)=f\left(\left(\frac{1-p}{2}\right)^2+p\right)=f\left(\left(\frac{1-p}{2}\right)^2\right)$$ From $(2)$, we have $$\frac{1+p}{2}f\left(\frac{1+p}{2}\right)=\frac{1-p}{2}f\left(\frac{1-p}{2}\right).$$ However $$f\left(\frac{1+p}{2}\right)=f\left(\frac{1-p}{2}+p\right)=f\left(\frac{1-p}{2}\right).$$ This means $$f\left(\frac{1+p}{2}\right)=0$$ for all $p\in P$ such that $p\ne 0$. Hence, for $p\in P\setminus\{0\}$, $\frac{1+p}{2}\in P$, so $1+p \in P$, making $1\in P$. This shows that either $P=\{0\}$ or $\mathbb{Z}\left[\frac12\right]\subseteq P$.
If $P=\{0\}$, then $f(x)-x=0$ for all $x\in \Bbb R$. Therefore, $f(x)=x$ for every $x\in \Bbb R$. This yields one solution of $(1)$. From now on $P\neq \{0\}$. Thus $\mathbb{Z}\left[\frac12\right]\subseteq P$. We want to show that $P=\Bbb R$, so that $f(x)=0$ for every $x\in \Bbb R$, and this is another solution of $(1)$.
For an arbitrary $y\in\mathbb{R}$, we see that $$f\left(\left(y+\frac{1}{2}\right)^2\right)=\left(y+\frac{1}{2}\right)f\left(y+\frac{1}{2}\right)=\left(y+\frac{1}{2}\right)f(y),$$ since $1/2\in P$. That is $$f\left(\left(y+\frac12\right)^2\right)=\frac{1}{2}f(y)+yf(y)=\frac12f(y)+f(y^2).$$ Consequently \begin{align}f\left(\left(y+\frac12\right)^2\right)-\left(y+\frac12\right)^2&=\frac{1}{2}f(y)+f(y^2)-\left(y+\frac12\right)^2\\&=\left(\frac{1}{2}f(y)-y\right)+\left(f(y^2)-y^2\right)+\frac14.\end{align} Because $f\left(\left(y+\frac12\right)^2\right)-\left(y+\frac12\right)^2$, $f(y^2)-y^2$, and $\frac14$ are all elements of $P$, we conclude that $$\frac{1}{2}f(y)-y\in P.$$ Therefore, $$\big(f(y)-y\big)-y=2\left(\frac{1}{2}f(y)-y\right)\in P.$$ As $f(y)-y\in P$, we get $y\in P$.