Incorrect manipulation of limits

The quotient property of limit says that if $\lim\limits_{x\to a}f(x)=L$ and $\lim\limits_{x\to a}g(x)=M\neq 0$, then $$\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M} = \frac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}g(x)}.$$ But this requires $M\neq 0$.

In your very first equality, you attempt to use this when $M=0$ (since $\lim\limits_{h\to 0}h = 0$). This is not a valid use of the limit laws/properties of limits.

The answers already posted fully answer your question. So what follows is not an answer to your question, but it may be helpful.

Let us assume that $f$ and $g$ are differentiable at $x$. Note that $$f(x+h)g(x) - f(x)g(x+h)= f(x+h)g(x)+(f(x)g(x)-f(x)g(x))-f(x)g(x+h).$$ We have added $0$ in the middle, which is harmless. A trick that looks very similar was undoubtedly used in your book or notes to prove the product rule for differentiation.

Rearranging a bit, and with some algebra, we find that $$f(x+h)g(x) - f(x)g(x+h)=(f(x+h)-f(x))g(x)-f(x)(g(x+h) -g(x)),$$ and therefore $$\frac{f(x+h)g(x) - f(x)g(x+h)}{h}=\frac{(f(x+h)-f(x))g(x)}{h}-\frac{f(x)(g(x+h) -g(x))}{h}.$$

The rest is up to you.

Added stuff, for the intuition: The following calculation is way too informal, but will tell you more about what's really going on than the mysterious trick.

When $h$ is close to $0$, $$f(x+h) \approx f(x)+hf'(x)$$ with the approximation error going to $0$ faster than $h$. Similarly, $$g(x+h) \approx g(x)+hg'(x).$$ Substitute these approximations into the top. Simplify. Something very pretty happens!

Hint to solve your problem: Let $$ \varphi (h) = \frac{{f(x + h)g(x) - f(x)g(x + h)}}{h}. $$ Then $$ \varphi (h) = \frac{{[f(x + h) - f(x) + f(x)]g(x) - f(x)[g(x + h) - g(x) + g(x)]}}{h}. $$ From this it follows straightforwardly that $$ \mathop {\lim }\limits_{h \to 0} \varphi (h) = g(x)f'(x) - f(x)g'(x). $$ I'll give more hints if you need.