For finding the derivative, $\frac{\mathrm{d}y}{\mathrm{d}x}$ of an implicit function, there is this formula,
When $f(x,y)=0$, $$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{\text{Differentiation of $f$ w.r.t to $x$ keeping $y$ constant}}{\text{Differentiation of $f$ w.r.t to $y$ keeping $x$ constant}}$$
I want to understand how this method came about.
Is there a clear proof for this formula?
It's pretty easy to show if you use partial differentials with a slightly different notation. Let's let $\partial_x f$ represent the partial differential of $f$ when we allow $x$ to freely vary (i.e., when $y$ is held constant). Note that when we allow $x$ to freely vary, then its differential is always the same as the total differential, so it can be notated as $dx$. Similarly with $y$. So, the partial differential of $f$ with respect to $x$ is $\frac{\partial_xf}{dx}$.
Now, let's put this to use. A total derivative is the sum of its partials. Therefore, the total derivative of $f$ is going to be $\partial_xf + \partial_yf$. Since the right-hand side is a constant, its derivative will be zero. Therefore: $$ \partial_xf + \partial_yf = 0 \\ \partial_xf = -\partial_yf \\ \frac{\partial_xf}{\partial_yf} = -1 $$
Now, what are our two partial derivatives? They are $\frac{\partial_xf}{dx}$ and $\frac{\partial_yf}{dy}$. What happens if we put them in ratio with each other? $$ \frac{\frac{\partial_xf}{dx}}{\frac{\partial_yf}{dy}} $$ Using fraction rules, we get (notice the "-1" comes from the equivalence above): $$ \frac{\frac{\partial_xf}{dx}}{\frac{\partial_yf}{dy}} = \frac{\partial_xf}{dx} \cdot \frac{dy}{\partial_yf} = \frac{\partial_xf}{\partial_yf}\cdot\frac{dy}{dx} = -1\cdot\frac{dy}{dx} = -\frac{dy}{dx} $$ This is what we were looking for. On the left is the ratio of the two partial derivatives. On the right is the total derivative with a negative sign.