Consider the following singular ODE $$ (t^{N1}g(y'))'=t^{N1}f(y) $$ with initial condition $$y(0)=y_0$$ $$y'(0)=0$$. where $g$ is and increasing function. How one can prove that this problem has a solution?

One cannot prove existence in the generality that you have stated it. For example, if $g(p) = e^p$, there is no solution when $N>1$ and $f$ is differentiable. To see this, observe that, if there were a solution $y(t)$ to your problem in a neighborhood of $t=0$, then the curve $(t,y,p) = (t,y(t),y'(t))$ would be an integral curve of the system $$ dy  p\ dt = d(t^{N1}g(p))  t^{N1}f(y)\ dt = 0 $$ that passes through the point $(t,y,p) = (0,y_0,0)$. Now, off the hypersurface $t=0$ in $typ$space, this is equivalent to the system $$ dy  p\ dt = \bigl((N{}1)g(p)tf(y)\bigr)\ dt + tg'(p)\ dp = 0. $$ As long as $g(0)\not=0$ and $N>1$, this system has rank $2$ at the point $(t,y,p) = (0,y_0,0)$, so that there is a unique integral curve of the system passing through this point. However, inspection shows that the curve is $(t,y,p) = (0,y_0,p)$ is an integral curve that passes through the point in question. By uniqueness, it is the only one. Thus, there is no integral curve of the kind that you would get from a solution to your problem. I suspect that you have left out some hypotheses or not made a correct symmetry reduction of the system that you are trying to study. 

