Computing the derivative from the left gives you
$$
\lim\limits_{h\rightarrow0^-} {f(0+h)-f(0)\over h }
=\lim\limits_{h\rightarrow0^-} {h-1\over h }=\infty.
$$
(In particular, note $f(0)=1$, not $0$.)
You can also see the derivative from the left doesn't exist (as a real number) by considering slopes of secant lines. Note a secant line has one endpoint at the point $(0,1)$ and the other at a point $(h,h)$ with $h<0$. As $h$ tends to $0$, the slopes tend to $\infty$.
The function defined by
$$f(x)=\begin{cases}\frac{\sin(x)}{x}&,x\ne 0\\\\1&,x=0\end{cases}$$
is not only differentiable at $x=0$, it is continuously differentiable there.
NOTE:
I thought it would be instructive to present a way forward that relies only on a standard, elementary inequality and the squeeze theorem. To that end, we proceed.
The derivative at $x=0$ is given by
$$f'(0)\equiv \lim_{h\to 0}\frac{\frac{\sin(h)}{h}-1}{h} \tag 1$$
Recalling from elementary geometry that the sine function satisfies the inequalities
$$\cos(h) \le \frac{\sin(h)}{h}\le 1 \tag 2$$
for $|h|\le \pi/2$, we see that the term under the limit in $(1)$ satisfies the inequalities
$$-2\sin^2(h/2)= \cos(h)-1\le \frac{\sin(h)}{h}-1\le 0 \tag 3$$
Then, taking absolute values, dividing by $|h|$, and using the right-hand side inequality in $(2)$ yields
$$0 \le \left|\frac{\frac{\sin(h)}{h}-1}{h}\right|\le \frac12 |h| \tag 4$$
whereupon applying the squeeze theorem to $(4)$ produces the limit
$$\lim_{h\to 0}\frac{\frac{\sin(h)}{h}-1}{h}=0$$
Therefore, $f'(0)=0$.
For $x\ne 0$, we have
$$f'(x)=\frac{x\cos(x)-\sin(x)}{x^2}$$
To see that $f'(x)$ is continuous at $x=0$, we need to show that
$$\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2}=0 $$
Again, using $(2)$ we see that
$$0\le \left|\frac{x\cos(x)-\sin(x)}{x^2}\right|\le\left|\frac{1-\cos(x)}{x}\right| \le \frac12 |x| \tag 5$$
whereupon applying the squeeze theorem to $(5)$ produces the limit
$$\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2}=0 $$
Therefore,
$$\lim_{x\to 0}f'(x)=0=f'(0)$$
which shows that $f'$ is continuously differentiable at $0$.
Best Answer
Your work is correct. Notice that the limit on the right and on the left are the derivative of $1-e^{-x}$ and $\ln(1-x)$ respectively at $0$ so you need to verify that they are equal.