Check my workings: Show that $\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$

Try applying L'Hospital's Rule to $h$, that is, differentiate with respect to $h$.

When you use that $f'(x) = \lim_{h \to 0} \frac {f(x+h)-f(x-h)}{2h}$, you get that $f''(x) = \lim_{k \to 0} \lim_{h \to 0} \frac {f(x+h+k)-f(x+h-k)-f(x-h+k)+f(x-h-k)}{4hk}$.

Let $g_x(h,k)$ be that expression : $f''(x) = \lim_{k \to 0} \lim_{h \to 0} g_x(h,k)$, while what you are given is $\lim_{h \to 0} g_x(h/2,h/2)$.
Both limits are going to $(0,0)$ but not along the same path. So it makes sense that they should be equal under certain conditions, namely if the function $g_x$ can be continuously extended at the point $(h=0,k=0)$ and on the two axis $h=0$, $k=0$: then no matter what path you take in your limit to $(0,0)$, you will get the same result.

In order to show that it is the case, you need to use the fact that $f''$ is continuous. Apply the mean value theorem to the functions $g_{x,h} : k \mapsto f(x+h+k) - f(x-h+k)$ : forall $h,k$, there is a $k'$ such that $|k'|\le |k|$ and $g_{x,h}(k) - g_{x,h}(-k) = 2kg'_{x,h}(k')$, which means that $g_x(h,k)$ simplifies to $\frac{g'_{x,h}(k')}{2h} = \frac{f'(x+h+k') - f'(x-h+k')}{2h}$.
Note that with the continuity of $f'$, this implies that for $h \neq 0$, $g_x$ can be continuously extended at $g_x(h,0)$ by $g_x(h,0) = \frac {f'(x+h)-f'(x-h)}{2h}$ (and similarly on the other axis)

Next we can apply the mean value theorem again, to all the functions $h \mapsto f'(x+h+k')$:
forall $h,k$ there are some $h',k'$ such that $|h'| \le |h|, |k'| \le |k|$, and $g_x(h,k) = f''(x+h'+k')$.

Then, we use the continuity of $f''$ to conclude that $\lim_{(h,k) \to (0,0)} g_x(h,k) = \lim_{(h,k) \to (0,0)} f''(x+h'+k') = f''(x)$

Apply Taylor's formula in the form $f(x+h) = f(x) + h f'(x) + h^2 f''(x)/2 + o(h^2 f''(x))$.