When you write down an equation in $x$ and $y$ of the form
$$F(x,y)=0,$$
then you can define a curve $\mathcal{C}$ in the plane by saying that $(x_0,y_0)\in\mathcal{C}$ if and only if $F(x_0,y_0)=0$... a point is on the curve if and only if it satisfies the equation of the curve.
Now for different $F(x,y)$ you get different curves. Some might be empty such as $x^2+y^2+1$ while others are the entire plane such as $\sin^2x+\cos^2x-1$.
Others are nicer such as $x^2+y^2-1$ which gives you a circle or the equation that gives a figure-of-eight. Now locally (i.e. near a specific point) these curves might not be that nice in that, no, they don't look the graphs of functions or differentiable functions.
For example, the point in the middle of the figure-of-eight does not look like the graph of a function. Similarly on the left- and right-hand sides of the circle there are vertical tangents --- this is not the behaviour of a differentiable function.
However away from such danger points --- such as anywhere else on the circle --- if you zoom in close enough to the point, locally, the circle looks like the graph of a function. So what we do is almost take each point $(x,y)$ on a case-by-case basis and say, O.K., near this point we could in theory define a function $y=y(x)$.
For the example of your circle. Suppose we are away from the vertical tangents and are thinking about a point $P=(x,y)$. The implicit function tells you that close to $P$ there is a function $y=y(x)$ whose graph is locally the same as the circle (in fact for the circle you have $y(x)=\pm\sqrt{1-x^2}$).
So we assume that, near $P$, we actually have (I actually get my students to write out $y=y(x)$ to help them see Chain Rules and not have $y'=1$)
$$x^2+[y(x)]^2-1=0.$$
Now these are two functions (LHS and RHS) so have the same derivative:
$$2x+2(y(x))\cdot\frac{dy}{dx}=0\Rightarrow x+y\frac{dy}{dx}=0...$$
Taking the derivative of $\displaystyle\frac{y^2}{36}$, we have
$$\frac{d}{dx}\left(\frac{y^2}{36}\right)=\frac{d}{dx}\left(\frac{1}{36}*y^2\right)=\frac{1}{36}*\frac{d}{dx}\left(y^2\right)=\frac{1}{36}*2y\frac{d}{dx}\left(y\right)=\frac{2y}{36}*\frac{dy}{dx}=\frac{y}{18}*\frac{dy}{dx}.$$
In the step where we take the derivative of $y$, since $y$ is a function in terms of $x$, it's derivative $\displaystyle\frac{dy}{dx}$ represents the change in $y$ with respect to $x$.
To answer your question, as to why the second term in your WolframpAlpha query is simply $\displaystyle\frac{2x}{25}$, notice what $\displaystyle\frac{d}{dx}(x)=\frac{dx}{dx}$ represents: $\textbf{the change in x with respect to x}$, which is $1$.
So, taking the derivative of $\displaystyle\frac{x^2}{25}$, we have
$$\frac{d}{dx}\left(\frac{x^2}{25}\right)=\frac{1}{25}*\frac{d}{dx}\left(x^2\right)=\frac{1}{25}*2x\frac{d}{dx}\left(x\right)=\frac{2x}{25}*1=\frac{2x}{25}.$$
Best Answer
It doesn't make any more sense to "prove implicit differentiation" than it does to "prove numbers," but I assume you're asking why implicit differentiation is valid i.e. preserves the truth of equations.
Implicit differentiation is just an application of the chain and other derivative rules to both sides of an equation, with (in the usual case) $y$ an abridgment of $f(x)$. Observe:
$$\frac{d}{dx}g(f(x),x)=\frac{\partial g}{\partial f}(f(x),x)\cdot\frac{df}{dx}(x)+\frac{\partial g}{\partial x} \tag{1}$$
$$\left(g(y,x)\right)'=\frac{\partial g}{\partial y}y'+\frac{\partial g}{\partial x} \tag{2}$$
The first is how you would take the derivative of a function of both $y=f(x)$ and $x$, but notice the difference between $d/dy$ and $\partial/\partial y$, while the latter equation is the same thing in fewer symbols courtesy of implicit differentiation. You always use ID in contexts where you have a function of both dependent and independent variables, and it's valid because of the above analogy - and it indeed generalizes to arbitrarily many dependent (or independent!) variables. For example, if we understand $u$ and $v$ to be functions of $t$ then we can differentiate
$$u^2+tv^2=1\quad\longrightarrow\quad 2uu'+(1\cdot v^2+t\cdot2vv')=0.$$