Let $f(z)=\Re(z)\in\mathbb{C}$, $z=x+iy$ and $h=h_x+ih_y$, then
$$f'(z)=\lim_{h\to0}\frac{f(z+h)-f(z)}{h}$$ $$= \lim_{h\to0}\frac{f(x+h_x+i(y+h_y))-f(x+iy)}{h}$$ $$= \lim_{h\to0}\frac{\Re(x+h_x+i(y+h_y))-\Re(x+iy)}{h}$$ $$= \lim_{h\to0}\frac{x+h_x-x}{h}$$ $$= \lim_{h\to0}\frac{\Re(h)}{h}$$ now, as we take the limit, it's worth considering how our expression behaves depending on where we take the limit from. If we consider what happens as we take the limit as $h\to0$ along the real axis, we have that $\Re(h)=h$, hence the ratio between $\Re(h)$ and $h$ is $1$, and the limit hence is $1$. However, if we repeat this process, but along the imaginary axis, we find that $\Re(h)=0$, and hence the ratio between $\Re(h)$ and $h$ is $0$, and our limit evaluates to $0$. We notice that already, the limit approaches different values as $h\to0$.
My question, then, is whether or not it is therefore safe to say that the limit of the difference quotient does not exist, and that therefore, $f(z)$ is not differentiable, or are there other conditions that we need to check before we jump to such a conclusion?
You correctly demonstrate this, you prove that the limit does not exist. Another way of prove this is using the so-called Cauchy-Riemman equations, it is easily proof that if a funcion is holomorphic if and only if the real part function and the imaginary part function are differentiable as funcions on $\mathbb{R}²$ and the Cauchy-Riemman equations holds, they are $u_x = v_y$ and $u_y = - v_x$. It follows that if $f: \Omega \subset \mathbb{C} \to \mathbb{R}$ is holomorphic then $v_y = 0$ and $v_x = 0$ then $f$ is constant(Assuming $\Omega$ is conected).