If $f'(t)$ exists then $\lim_{h,k\to0}\frac{f(t+h)-f(t-k)}{h+k}=f'(t)$ Consider $f'$ continuous...
I'm tried to prove it...with Taylor's expansion:
$f(t+h)=f(t)+hf'(t)+h\phi(t,h)$
$f(t-k)=f(t)-kf'(t)-k\phi(t,-k)$
Then:
$\frac{f(t+h)-f(t-k)}{h+k}=f'(t)+\frac{h}{h+k}\phi(t,h)+\frac{k}{h+k}\phi(t,-k)$
I don't know how I can prove $\lim_{h,k\to0}\frac{h}{h+k}\phi(t,h)=0$

You went wrong way. It is simpler:
$$\frac{f(t+h)-f(t-k)}{h+k}=\frac{f(t+h)-f(t)+f(t)-f(t-k)}{h+k}=$$
$$h\frac{f(t+h)-f(t)}{h(h+k)}+k\frac{f(t)-f(t-k)}{k(h+k)}$$
And now
$$\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}=\lim_{k\to 0}\frac{f(t)-f(t-k)}{k}=f'(t)$$
making
$$\lim_{h,k\to 0} \left ( h \frac{f(t+h)-f(t)}{h(h+k)}+k\frac{f(t)-f(t-k)}{k(h+k)} \right ) =$$
$$f'(t)\lim_{h,k\to 0}\left ( \frac{h}{h+k}+\frac{k}{h+k} \right ) =f'(t)\lim_{h,k\to 0}\frac{h+k}{h+k}=f'(t)$$
Clarification:
To be totally precise, even for untrained eyes, this should be derived in the opposite direction.
If $\lim a$ exists and $\lim b$ exists then $\lim ab$ exists.
$$\lim_{h,k\to 0}\frac{h+k}{h+k}=1$$
$$\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}=\lim_{k\to 0}\frac{f(t)-f(t-k)}{k}=f'(t)$$
So
$$\lim_{h,k\to 0}f'(t)\frac{h+k}{h+k}=f'(t)$$
$$\lim_{h,k\to 0} \left ( \frac{hf'(t)}{h(h+k)}+\frac{kf'(t)}{k(h+k)} \right ) = f'(t)$$
and now expand $f'(t)$ in two different but equivalent ways and there you go.
The entire trick here is dealing with expressions that have predefined limits.
