I know that if $f$ is Lipschitz or $C^1$ then the IVP $x'=f(t,x), x(t_0)=x_0$ has a solution, which is also unique.
Now I'm wondering whether the IVP $y'(x)=h(x)g(y(x)), y(x_0)=y_0$ ($h\in C([x_0-h,x_0+h],\mathbb{R}), g\in C([y_0-\delta,y_0+\delta],\mathbb{R}), g(y_0)\neq 0$) has a solution which is also unique.
Attempt 1: I considered $\Omega:=[x_0-h,x_0+h]\times [y_0-\delta,y_0+\delta]$ and $f\in C(\Omega,R),\ f(x,y):=h(x)g(y(x))$ which would guarantee a unique solution by Peano-Picard theorem if $f$ were Lipschitz or $C^1$ which is not necessarily the case unfortunately.
Attempt 2: I tried by separating variables, obtaining $\int \frac{dy}{g(y(x))}=\int h(x)dx$ but not knowing the expression of $g$ and $h$ I don't see how I can go further.
I'm stuck so I'd appreciate some help.
Thanks.