The Dirac-delta function has the property, $$ x \delta'(x) = -\delta(x) $$
If I consider an ODE of the following form, $$ \frac{df}{dx} = \frac{-1}{x} f(x) $$ here we can show that $f(x)$ satisfies the above property and therefore $f(x) = \delta(x)$. However, if I solve the above ODE using separability, $$ \int \frac{df}{f} = -\int \frac{dx}{x} $$ I get a result of the kind $f(x) \sim 1/x$. How do I reconcile between these two seemingly contradictory solutions.
Some obstacles to your wish:
The solution of an ODE is always a continuously differentiable function.
$ xf'(x)+f(x)=0 $ is not defined as ODE at $x=0$ (but to both sides of it). So how any generalized solution behaves at $x=0$ is not covered by ODE theory.
There are different ways how one can try to extend a solution to $x=0$, but most will involve that the extension be at least continuous.
One can try to modify the equation slightly so that the new equation is regular, such as $$ (x^2+\varepsilon^2)f'(x)+xf(x)=0\implies f(x)=\frac{c}{\sqrt{x^2+ε^2}}. $$ But this solution does not have a finite area under the curve, so it can not be used to construct an approximation of the Dirac delta.