This limit is immediate using L'Hopital rule:
$$ \lim_{h\rightarrow 0} \frac{f\big(x + h(\alpha - x)\big) - f(x)}{h}= \lim_{h\rightarrow 0} f'\big(x + h(\alpha - x)\big) (\alpha - x)= f'(x) (\alpha - x)\\ $$
Anyway I wonder if it can be equally simple to solve it without L'Hopital.
By the MVT $$\frac{f\big(x + h(\alpha - x)\big) - f(x)}{h}=f'(x_h)(\alpha - x)$$ with $x_h$ between $x$ and $x+h(\alpha - x)$.
In the second equality, you are supposing thaf $f'$ is continuous. With this hypothesis: $$ \lim_{h\to 0} \frac{f\big(x + h(\alpha - x)\big) - f(x)}{h}= \lim_{h\to 0} f'(x_h)(\alpha - x)=f'(\lim_{h\to 0} x_h)(\alpha - x)= f'(x)(\alpha - x). $$