What is the partial derivative of this equation in terms of x I want to calculate $$F(x)=\frac{\partial {\ln (f(x,g(x)))}}{\partial{x}}$$
should the answer be $$\frac{1}{f(x,g(x))}\frac{\partial f(x,g(x))}{\partial x}\frac{dg(x))}{dx}$$ 
 A: First note $$H(x)= \ln f(x,g(x)).$$ You're using partial derivatives. However in your case, it is not necessary as $H$ only depends on $x$. The derivative of $H$ is
$$\frac{\left( f(x,g(x)) \right)^\prime}{f(x,g(x))}$$.
So you have now to compute the derivative of $x \mapsto f(x,g(x))$ which is $$\frac{\partial f}{\partial x}(x, g(x)) + g^\prime(x) \frac{\partial f}{\partial y} (x,g(x))$$
Finally
$$F(x)=\frac{\frac{\partial f}{\partial x}(x, g(x)) + g^\prime(x) \frac{\partial f}{\partial y} (x,g(x))}{f(x,g(x))}$$
A: I think you are missing a derivative there.
To see that it might be helpful to consider your problem from the viewpoint of multivariate analysis.
Take $f:R^2 \to R, (x,y) \mapsto f(x,y)$ and $h: R \to R^2, x \mapsto (x,g(x))^T$. 
You are now looking for $d/dx (f \circ h)(x)$.
Using the (multivariate) chain rule (see https://en.wikipedia.org/wiki/Chain_rule#Higher_dimensions) this will be given as the product of the associated Jacobian matrices leading to ($g'(x)$ is the shrothand for $d/dx g(x)$)
\begin{align}
\frac{d}{dx}(f \circ h)(x) = \left(\frac{\partial}{\partial x}f(x,y)|_{(x,g(x))}, \frac{\partial}{\partial y}f(x,y)|_{(x,g(x))}\right) \left(1, g'(x)\right)^T\\
=\frac{\partial}{\partial x}f(x,y)|_{(x,g(x))} + \frac{\partial}{\partial y}f(x,y)|_{(x,g(x))}g'(x).
\end{align}
The logarithm will indeed have the usual effect, i.e. you finally have
$$
\frac{d}{dx}\log((f \circ h)(x)) = \frac{ \frac{\partial}{\partial x}f(x,y)|_{(x,g(x))} + \frac{\partial}{\partial y}f(x,y)|_{(x,g(x))}g'(x)}{f(x,g(x))}
$$
