If the domain of $f$ in $f'(x)=-f(x^2)$ is $(1,\infty)$ as given in the linked AOPS problem, then we can find a solution which is not of the form of $c/x$ for constant $c$.
Note that if $f(x)=c/x$, then necessarily $c=2f(2)$ by putting $x=2$. With this choice of $f$, we have $F(x)=f(x)-\frac{2f(2)}x = 0$.
For any $f$ satisfying $f'(x)=-f(x^2)$, we have $F(x)=f(x)-\frac{2f(2)}x$ also satisfies $F'(x)=-F(x^2)$ and we have $F(2)=0$. By constructing a nonzero function $F$ with
$$F'(x)=-F(x^2), \ \ F(2)=0. \ \ (1)$$
we see that not every function with $f'(x)=-f(x^2)$ is of the form $f(x)=c/x$.
Define $F(x)$ on $[2,4]$ the bump function on the interval $[2,4]$. Then we define $F(x)$ on $[4,16]$ by $F(x)=-F'(\sqrt x)$. The bump function is infinitely differentiable and we need $F'(x)$ on $[2,4]$ to define $F(x)$ on $[4,16]$. We repeat this procedure to define $F(x)$ on $[16,256]$, $[256, 65536]$, $\ldots$, in general $[2^{2^k}, 2^{2^{k+1}}]$ for $k\geq 0$. Then $F(x)$ satisfies $(1)$ for all $x\geq 2$.
To define the function on the interval $(1,2)$, we use
$$
F(x)=\int_x^2 F(t^2)dt.
$$
We need $F(x)$ on $[2,4]$ to define $F(x)$ on $[\sqrt 2, 2]$ by the above formula. To define $F(x)$ on $[\sqrt[4]{2},\sqrt 2]$ we need $F(x)$ on $[\sqrt 2, 4]$. Repeating the procedure, we are able to define $F(x)$ on $[2^{2^k}, 2^{2^{k+1}}]$ for any negative integer $k$. Thus, $F(x)$ is defined for all $x\in (1,2)$.
Combining these together, we have a nonzero function $F$ satisfying $(1)$ on $(1,\infty)$.
Best Answer
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ s.t:
$$f(x\cdot f(y))= \frac{f(x)}{y},\forall x,y \in \mathbb{R},y≠0$$ For $x=0$ we get $f(0)=0$ like you said.
For $x=y=1$ we get: $$f(1f(1))=\frac{f(1)}{1}\Rightarrow f(f(1))=f(1)$$
Now, based on what we have to prove, we can't have $f(1)=0$ since then, if we let $y=1$ we'd get $\forall x\in\mathbb{R}$:
$$f(xf(1))=f(x)\Rightarrow f(0)=f(x)\Rightarrow f(x)=0\Rightarrow f(x)f(1/x)=0\neq1,\forall x\in\mathbb{R}^{*}.$$
So $f(1)\neq 0$ and hence we can set $y=f(1),x=1$ to get:
$$f(f(1))=\frac{f(1)}{f(1)}\Rightarrow f(f(1))=1$$ and since $f(f(1))=f(1)$ we have $f(1)=1$.
Next, let $x=1,y\neq 0$ to get:
\begin{equation} \tag{1} f(f(y))=\frac{f(1)}{y}\Rightarrow yf(f(y))=1 \end{equation} Now let $x=1,y=\frac1y$. Then:
\begin{equation} \tag{2} f\left(f\left(\frac1y\right)\right)= yf(1)=y \end{equation}
Finally, setting to $(1)$ $y = f\left(\frac1y\right)$ we get:
\begin{equation} \tag{3} f\left(\frac1y\right)f\left(f\left(f\left(\frac1y\right)\right)\right)=1 \end{equation}
and since from $(2)$ we have $f\left(f\left(\frac1y\right)\right)=y$, we conclude:
$$ (3)\Rightarrow f\left(\frac1y\right)f\left(y\right)=1$$.
$\textbf{Edit:}$ I should also justify the step $y=f\left(\frac{1}{y}\right)$, since in order for it to be valid, we must have $f\left(\frac{1}{y}\right)\neq 0,\forall y\in\mathbb{R}^{*}$ or equivalently $f(y)\neq 0,\forall y\in\mathbb{R}^{*}$.
If we had some $y_{0}\in\mathbb{R}^{*}$ such that $f(y_{0})=0$, then for all $x\in\mathbb{R}^{*}$ we'd get:
$$f(xf(y_{0}))=\frac{f(x)}{y_0}\Rightarrow f(0)=\frac{f(x)}{y_0}\Rightarrow 0 = f(x)$$ which can't be the case for the same reason we can't have $f(1)=0$