How can I solve the differential equation $f'(x) + f\left( x^2\right) =0$? I was looking at this AOPS thread which dealt with the following question:

Find all functions $f:(1, \infty) \to \mathbb{R}$ that satisfy $$f(x) -f(y) = (y-x)f(xy)  $$

I attempted to solve (at least part of) this question, and in one of my attempts I encountered a path that seemed promising, but I couldn't manage to finish the solution. Here's what I did:

Rewriting the functional equation as
$$
\frac{f(x) - f(y)}{x-y} = -f(xy)
$$
the LHS is the slope of a secant line. This makes me want to take the limit $y\to x$ on both sides to get some derivatives. So assuming $f$ is differentiable on the domain of interest, this gives
$$
f'(x) = -f\left(x^2\right) \tag{1}
$$
as a differential equation whose solutions form a subset of differentiable solutions to our original functional equation.
As shown in the AOPS thread, the family of solutions to the original problem is indeed $f(x) = \frac{c}{x}$, with $c$ some constant, and this family of functions do satisfy that $\frac{\mathrm{d}}{\mathrm{d} x}\frac{c}{x} = -\frac{c}{x^2}$ as the differential equation suggests, so all seems good up to this point.

Even though using the ansatz $f(x) = \frac{c}{x}$ to verify the differential equation $(1)$ does work, I don't know how to solve the differential equation directly. I thought about introducing a substitution of the form $u(x) = x^2$, but then the LHS of $(1)$ would end up with things like $f'(\sqrt{u})$ which again doesn't seem very useful.
Is it possible to solve the differential equation $(1)$ directly? And if the answer is yes, what is the procedure with which you can go about solving a differential equation like this? Thank you!
 A: 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 $(1,\infty)$.
A: With $a_0 := 1$, $a_k := (-1)^k/\prod_{j=1}^k\, (2^j - 1)$ for $k > 0$, another linearly independent solution is
\begin{equation*}
g(x) := \sum_{k=0}^\infty\, a_k\,x^{2^k - 1}.
\end{equation*}
There's a note on this in OEIS A005329. To answer the 'how can I solve' part, try a Laurent series expansion.
