Summary: This solution shows that if a function $f:\mathbb{R}\to\mathbb{R}$ satisfies $f(1)=1$ and $f\left(yf(x)+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$ for all $x,y\in\mathbb{R},y\neq0$, then $f(0)=0$ and $f(x)=\frac{1}{x}$ for all $x\neq0$. No additional assumptions on $f$ are necessary!
Thanks @Sil for giving all these references! I meanwhile came up with a solution myself, and last but not least because not many solutions of this problem seem to be around, I would like to share mine.
Suppose $f$ is a solution to this functional equation and for $x,y\in\mathbb{R},\ y\neq0$ write $P(x,y)$ for the assertion $f\left(yf(x)+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$. Furthermore, we define $\mathcal{N}:=\{x>0\ |\ f(x)=0\}$, and assume that $\mathcal{N}\neq\emptyset$.
As @Kevin already pointed out, we have $\alpha\in\mathcal{N}\implies\alpha^2+1\in\mathcal{N}$, and in particular $\mathcal{N}$ is unbounded. Furthermore, $P(1,\alpha)$ gives also $\alpha\in\mathcal{N}\implies\alpha+\frac{1}{\alpha}\in\mathcal{N}$. Now, if $\alpha,\beta\in\mathcal{N}$ then
$$
P(\alpha,\beta):\quad f\left(\frac{\alpha}{\beta}\right)=\alpha\beta f\left(\alpha^2+\beta^2\right)\\
P(\beta,\alpha):\quad f\left(\frac{\beta}{\alpha}\right)=\beta\alpha f\left(\beta^2+\alpha^2\right)
$$
and thus $f\left(\frac{\alpha}{\beta}\right)=f\left(\frac{\beta}{\alpha}\right)$. Therefore, if $\alpha\in\mathcal{N}$ then
$$
f\left(\frac{1}{\alpha}\right)=f\left(\frac{\alpha+\frac{1}{\alpha}}{\alpha^2+1}\right)=f\left(\frac{\alpha^2+1}{\alpha+\frac{1}{\alpha}}\right)=f(\alpha)=0
$$
and thus also $\frac{1}{\alpha}\in\mathcal{N}$. This gives, together with the unboundedness of $\mathcal{N}$, the existence of $(\alpha_n)_{n\in\mathbb{N}}\in\mathcal{N}^\mathbb{N}$ with $\lim_{n\to\infty}\alpha_n=0$.
Now notice that for $x\neq 0$ and $\alpha\in\mathcal{N}$
$$
P(\alpha,\alpha^2 x):\quad f\left(\frac{1}{\alpha x}\right)=\alpha^3 x f\left(\alpha^4\left(x^2+\frac{1}{\alpha^2}\right)\right)\\
P(\frac{1}{\alpha}, x):\quad f\left(\frac{1}{\alpha x}\right)=\frac{x}{\alpha} f\left(x^2+\frac{1}{\alpha^2}\right)
$$
and thus $\alpha^4 f\left(\alpha^4\left(x^2+\frac{1}{\alpha^2}\right)\right)=f\left(x^2+\frac{1}{\alpha^2}\right)$ for all $x\neq 0$, or in more simple terms
$$
(*)\qquad\alpha^4 f(\alpha^4 z) = f(z) \quad\forall \alpha\in\mathcal{N},\ z>\frac{1}{\alpha^2}
$$
Now for a fixed $\alpha\in\mathcal{N}$ and $y>0$ let $n\in\mathbb{N}$ be such that $y>\max\{\frac{\alpha_n^2}{\alpha^4},\frac{\alpha_n^4}{\alpha^2},\alpha_n^2\}$, which exists as $(\alpha_n)\to 0$. Then
$$
\alpha^4 f(\alpha^4 y)=\frac{\alpha^4}{\alpha_n^4} \alpha_n^4 f\left(\alpha_n^4\frac{\alpha^4 y}{\alpha_n^4}\right)\overset{\frac{\alpha^4 y}{\alpha_n^4}>\frac{1}{\alpha_n^2}}{=}\frac{1}{\alpha_n^4}\alpha^4 f\left(\alpha^4\frac{y}{\alpha_n^4}\right)\overset{\frac{y}{\alpha_n^4}>\frac{1}{\alpha^2}}{=}\frac{1}{\alpha_n^4}f\left(\frac{y}{\alpha_n^4}\right)\overset{y>\alpha_n^2}{=}f(y).
$$
Thus we can strengthen $(*)$ and actually have
$$
(**)\qquad\alpha^4 f(\alpha^4 z) = f(z) \quad\forall \alpha\in\mathcal{N},\ z>0.
$$
In particular, for $z=\frac{1}{\alpha^2}$ we get $\alpha^4f(\alpha^2)=f\left(\frac{1}{\alpha^2}\right)$. On the other hand, we see that as $1+\alpha^2,1+\frac{1}{\alpha^2}\in\mathcal{N}$, we have by again combining $P(1+\alpha^2,1+\frac{1}{\alpha^2})$ and $P(1+\frac{1}{\alpha^2},1+\alpha^2)$ that
$$
f(\alpha^2)=f\left(\frac{1+\alpha^2}{1+\frac{1}{\alpha^2}}\right)=f\left(\frac{1+\frac{1}{\alpha^2}}{1+\alpha^2}\right)=f\left(\frac{1}{\alpha^2}\right)
$$
and thus, as $\alpha>0$ and $\alpha\neq 1$, we have $\alpha^4f(\alpha^2)=f\left(\frac{1}{\alpha^2}\right)=f(\alpha^2)\implies f(\alpha^2)=0$ so $\alpha^2\in\mathcal{N}$. But then also $\alpha^4\in\mathcal{N}$ which gives by $(**)$
$$
0=\alpha^4 f(\alpha^4)\overset{(**)}{=} f(1)=1,
$$
contradiction!
Therefore we conclude that $\mathcal{N}$ is empty, and as @Kevin already saw this gives $f(x)=\frac{1}{x}$ for all $x\neq 0$. As $P(0,y)$ gives $f(0)=0$, we have uniquely determined $f$, and we can verify easily that $f$ is indeed a solution of the equation at hand.