Prove this derivative is continuous

Let $\displaystyle U$ be a neighborhood of 0 in $\displaystyle \mathbb {R} $.

Suppose that $\displaystyle f:U \rightarrow \mathbb {R} $ is twice differentiable , f' and f'' are continuous, and $\displaystyle f(0)=0$.

Define $\displaystyle g: U \rightarrow \mathbb {R} $ by $\displaystyle g(x)= \frac {f(x)}{x} $ if $\displaystyle x \neq 0 $ and $\displaystyle \lim _{ h \rightarrow 0 } \frac {f(h)}{h} $ if $\displaystyle x = 0 $

Show that g' exists and is continuous on U.

Proof so far.

Claim: g' exists on U.

By the division theorem, we know that $\displaystyle f(x)=xg(x)$, implies that for $\displaystyle x \neq 0 $, we would have $\displaystyle f'(x)=xg'(x)+g(x)$, so we have $\displaystyle g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x} $.

For $\displaystyle x=0$, we have $\displaystyle g'(0)= \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h} - \frac {f(h)}{h}}{h} = 0 $

Claim: g' is continuous on U.

Write $\displaystyle f(x)=f(0)+f'(0)x+f'(t)x^2 \ \ \ \ \ 0<t<x$, so we have $\displaystyle f(x)=f'(0)x+f''(xh)x^2 \ \ \ \ \ 0<h<1$

In the case that $\displaystyle x \neq 0$:

Given $\displaystyle \epsilon > 0 $, pick $\displaystyle \delta > 0 $, then for each $\displaystyle x \in U $ and $\displaystyle |x-x_0|< \delta $, we have $\displaystyle |g'(x)-g'(x_0)|=| \frac {f'(x)- \frac {f(x)}{x}}{x} - \frac {f'(x_0)- \frac {f(x_0)}{x}}{x}|$$\displaystyle = \frac {|f'(x)-f'(x_0)|}{x} + \frac {| \frac {f(x)}{x}- \frac {f(x_0)}{x_0}|}{x}$

I have a feeling I'm very wrong as I'm stuck at this point, any hints? Thank you.