Hi everyone I'm asking two thinks is this proof correct (my other idea was using limit of sequences)? and are there a simpler alternative than this using Newton's approximation? If someone could help me I'd be so thankful.
Proposition: Let $X,Y$ be subsets of $\mathbb{R}$, $f:X\rightarrow Y$ and $g:Y\rightarrow \mathbb{R}$ be functions, $x_0\in X$, $y_0\in Y$ are limit points of $X$ and $Y$ respectively, and $y_0=f(x_0)$. Then if $f$ is differentiable at $x_0$ and $g$ is differentiable at $y_0$, then the composition is differentiable at $x_0$ and $(g\circ f)'(x_0)=g'(y_0)f'(x_0)$, i.e.,
$$\lim_{x\rightarrow x_0;x\in X\backslash\{x_0\}} \frac{(g\circ f)(x)-(g\circ f)(x_0)}{x-x_0}=g'(y_0)f'(x_0)$$
Proof: It will suffice to show that given a $\varepsilon>0$ there exists a $\delta>0$ such that $(g\circ f)(x)$ is $\varepsilon |x-x_0|$-close to $(g\circ f)(x_0)+g'(y_0)f'(x_0)(x-x_0)$ whenever $x\in X$ and $x$ is $\delta$-close to $x_0$. After a little algebra this is equivalent to
$g(f(x))-g(f(x_0))-g'(y_0)f'(x_0)(x-x_0)=$ \begin{align} \big[g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))\big] +\big[ g'(f(x_0))(\,(f(x)-f(x_0)-f'(x_0)(x-x_0))\,) \big]\end{align}
Since $\big|f(x)-f(x_0)-f'(x_0)(x-x_0) \big|\le |x-x_0| \varepsilon_1$ because $f$ is differentiable at $x_0$ by hypothesis. We set $\varepsilon_1=\varepsilon/2(| g'(f(x_0))|+1)$ and $\delta_1>0$ such that for all $\{x\in X: |x-x_0|\le \delta_1\}$ we have
$$\big|f(x)-f(x_0)-f'(x_0)(x-x_0) \big|\le \frac{ |x-x_0| \varepsilon}{2(| g'(f(x_0))|+1)}$$
Now let $\varepsilon_2=\varepsilon/2(|f'(x_0)|+1))$. Since $|g(y)-g(y_0)-g'(y_0)(y-y_0)|\le |y -y_0|\varepsilon_2$ we can find some $r>0$ such that whenever $|y-y_0|\le r$ the above inequality holds. Since $f$ is differentiable it follows that is continuous at $x_0$ then there is $\delta_2$, $0<\delta_2\le \delta_1$, such that $|f(x)-f(x_0)|\le r$ and $|f(x)-f(x_0)|\le |x-x_0|(|f'(x_0)|+1)$ at the same time. Thus we have
\begin{align}|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|\le \frac{|f(x) -f(x_0)|\varepsilon}{2(|f'(x_0)|+1)}\\ \le|x-x_0|\varepsilon/2 \end{align}
For instance we have
$$\begin{align}|g(f(x))-g(f(x_0))-g'(y_0)f'(x_0)(x-x_0)|\le |x-x_0|\varepsilon /2+ |g'(f(x_0))|\frac{ |x-x_0| \varepsilon}{2(| g'(f(x_0))|+1)} \\ \le |x-x_0| \varepsilon\end{align}$$
whenever $x$ is $\delta_2$-close to $x_0$ as claimed.
An idea for a much shorter and simpler proof using continuity, and using your same notation:
$$g\circ f(x_0):=\lim_{x\to x_0}\frac{g\circ f(x)-g\circ f(x_0)}{x-x_0}=$$
$$=\lim_{x\to x_0}\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}\cdot\frac{f(x)-f(x_0)}{x-x_0}=g'(f(x_0))\cdot f'(x_0)$$
since, by continuity, we have that
$$x\to x_0\implies f(x)\to f(x_0)$$