You're presumably given a function $f$ whose domain is a subset of $\mathbb R^2$. Since this isn't specified, I'll take the simpler approach and assume the domain is $\mathbb R^2$. This allow us to write $f\colon \mathbb R^2 \to \mathbb R$.
Also presumably, you're given a differential function $g\colon \mathbb R\to \mathbb R$.
Now let's define $\varphi$ as $\varphi \colon \mathbb R\to \mathbb R^2, x\mapsto (x, g(x))$.
You want to differentiate the function given below (with the appropriate implicit domain)
$$
x\mapsto \frac{\partial f}{\partial x}(x, g(x)) + \frac{\partial f}{\partial y}(x, g(x))g'(x)
$$
This can be rewritten as
$$
x\mapsto \left(\frac{\partial f}{\partial x}\circ\varphi\right)(x) + \left(\frac{\partial f}{\partial y}\circ\varphi\right)(x)\cdot g'(x)
$$
Since the derivative of a sum is the sum of the derivatives (when they all exist), we can focus on each member of this sum separately.
We want to differentiate $\color{blue}{x\mapsto \left(\dfrac{\partial f}{\partial x}\circ\varphi\right)(x)}$. Now allow me to change the notation $\dfrac{\partial f}{\partial x}$ to $\partial_1f$. They denote the same thing: the derivative with respect to the first coordinate.
In the notation of this answer we have $m=2$, $n=1=p$, $G=\partial_1f$, $F=\varphi$, (now because the choice of letters in both questions gets confusing, I'll use $\varphi_1$ and $\varphi_2$ (in place of $f_1$ and $f_2$ used in the linked answer), $\varphi_1\colon \mathbb R\to \mathbb R, x\mapsto x$ and $\varphi_2\colon \mathbb R\to \mathbb R, x\mapsto g(x)$. Now set $H=G\circ F$.
We have that $\color{blue}{H}$ is $\color{blue}{\left(\partial_1f\right)\circ \varphi}$ and this is what we wish to differentiate, in other words we wish to find $H'$. Since $H$ is a scalar function whose domain is $\mathbb R$, its components are just a singular $h_1$ in the notation of the linked answer, which yields:
$$
\begin{align}
\overbrace{
\begin{bmatrix}
\partial _1h_1
\end{bmatrix}_{x}
}^{H'(x)}
&=
\begin{bmatrix}
\partial_1\left(\partial_1f\right) & \partial_2\left(\partial_1f\right)
\end{bmatrix}_{\varphi(x,y)}
\overbrace{\begin{bmatrix}
\partial _1\varphi_1\\
\partial_1\varphi_2
\end{bmatrix}_{x}}^{\begin{bmatrix}
\varphi_1'\\
\varphi_2'
\end{bmatrix}_{x}}\\
&=
\begin{bmatrix}
\left(\partial _1\left(\partial_1f\right)\right)(x,g(x)) & \left(\partial_2\left(\partial_1f\right)\right)(x, g(x))
\end{bmatrix}
\begin{bmatrix}
1\\
g'(x)
\end{bmatrix}\\
&=
\begin{bmatrix}
\left(\partial _1\left(\partial_1f\right)\right)(x,g(x)) + \left(\partial_2\left(\partial_1f\right)\right)(x, g(x))\cdot g'(x)
\end{bmatrix}_.
\end{align}
$$
The last term above can be translated back to
$$
\dfrac{\partial^2f}{\partial x^2}(x,g(x)) + \dfrac{\partial^2f}{\partial y\partial x}(x, g(x))\cdot g'(x)
$$
It is similar for $\dfrac{\partial f}{\partial y}\circ\varphi$.
Best Answer
Your answers are fine. The chain rule gives us $$\frac{\partial}{\partial x} f(g(x,y)) = f'(g(x,y))\cdot \frac{\partial g}{\partial x} $$The only real work is the second piece. In calculating $\frac{\partial g}{\partial x}$ we treat anything which does not depend on $x$ as a constant. So $$\frac{\partial }{\partial x} \frac{x}{y} = \frac{1}{y}$$
In the same way that
$$\frac{d }{d x} \frac{x}{2} = \frac{1}{2}$$