I am not exactly sure what you are asking but as

**Aryth** said it seems to use the

implicit function theorem. In its most basic form it says that if $\displaystyle F:\mathbb{R}^2\mapsto \mathbb{R}$ (this notation it means that $\displaystyle F$ is a function of two variables) is a $\displaystyle \mathcal{C}^1$ function (ignore what this mean - it just means the function is behaved well-enough such as being differenciable). Say that $\displaystyle F(a,b) = 0$. The implicit function theorem says that if $\displaystyle \partial_y F(a,b) \not = 0$ then the equation $\displaystyle F(x,y) = 0$ can be solved

__uniquely__ (for $\displaystyle y$) in the "neighborhood" of $\displaystyle (a,b)$. The term "in the neighborhood" means in some small disk around $\displaystyle (a,b)$ we can solve this equation uniquely for $\displaystyle y$ (in terms of $\displaystyle x$).

For example, consider the circle $\displaystyle x^2+y^2 - 1= 0$. Where can we solve for $\displaystyle y$ uniquely (in a neighborhood)? Define the function $\displaystyle F(x,y) = x^2+y^2 - 1$. The point $\displaystyle (1,0)$ lies on this circle because $\displaystyle F(1,0) = 1^2 + 0^2 - 1 = 0$. If we try to solve the equation within the small disk $\displaystyle \sqrt{(x-1)^2+y^2} < \epsilon $ we see that it has two solutions, one in the upper half and one in the lower half (see red circle) - in fact those solutions are $\displaystyle y = \pm \sqrt{ 1-x^2}$. And this happens no matter how small we make the circle. Therefore, the equation $\displaystyle F(x,y) = 0$

**cannot** be solved

__uniquely__. Let us see what happens when we use the implicit function theorem. The theorem says that if $\displaystyle \partial_y F(1,0) \not = 0$ then we can solve the equation uniquely. Since we cannot solve this equation uniquely it must mean that $\displaystyle \partial_y F(1,0) = 0$. Check this: $\displaystyle \partial_y F(x,y) = 2y$ and so $\displaystyle \partial_y F(1,0) = 0$. However, if $\displaystyle F(a,b) = 0$ and $\displaystyle b\not = 0$ i.e. we stay away from the break in the graph where it lies above and below the x-axis then $\displaystyle \partial_y F(a,b) = 2b \not = 0$ and therefore we can solve this equation

__uniquely__. As in the case of the blue circle, the solution is $\displaystyle y=\sqrt{1-x^2}$.

Let us say that $\displaystyle F(x,y)$ is function which satisfies the condition of the implicit function theorem as in the first paragraph. Then in the neighborhood of $\displaystyle (a,b)$ (where $\displaystyle (a,b)$ is a solution to $\displaystyle F(x,y)=0$) we can solve for $\displaystyle y$

__uniquely__ in terms of $\displaystyle x$. This means we can define a new function $\displaystyle y = g(x)$ in this neighborhood. Remember a function is a set of pairs so that for each first coordinate there is a

__unique__ matching coordinate. Therefore, if we let $\displaystyle x$ be the first coordinate then by the theorem there is a

__unique__ matching coordinate so that $\displaystyle F(x,y) = 0$. And this would define a function $\displaystyle y=g(x)$.

But the implicit function theorem does not stop here, it says more it says that $\displaystyle g(x)$ is itself $\displaystyle \mathcal{C}^1$ (which is differenciable with some more properties). Now since $\displaystyle F(x,y) = 0 \implies F(x,g(x)) = 0$. The functions $\displaystyle F$ and $\displaystyle g$ are differenciable and so by the

chain rule for multivariable functions we get that (differenciating both sides of $\displaystyle F(x,g(x)) = 0$) $\displaystyle \partial_x F + \partial_y F \cdot g'(x) = 0$ and this means $\displaystyle g'(x) = - \frac{\partial_x F}{\partial_y F}$ (the partials are evaluated at $\displaystyle (x,y)$). And that gives you a formula.

If you got that then great. There is just one more point to be addressed. If it confuses you just ignore that. The questionable step is dividing by $\displaystyle \partial_y F$. How do we know it is non-zero? This is where we use two facts. The first one is that $\displaystyle \partial_y F (a,b) \not = 0$. The second one is that $\displaystyle F(x,y)$ is $\displaystyle \mathcal{C}^1$. The first fact is not good enough to say that $\displaystyle \partial_y F\not =0$ for all points around $\displaystyle (a,b)$ because it might be non-zero at $\displaystyle (a,b)$ and yet be zero at some points around $\displaystyle (a,b)$. This is where we need the second fact. The meaning of $\displaystyle \mathcal{C}^1$ is that the function is differenciable

**and** the derivative is continous. Therefore $\displaystyle \partial_y F$ is a continous function. Since $\displaystyle \partial_y (a,b)\not = 0$ it means there is a small enough neighborhood so that $\displaystyle \partial_y (x,y)\not = 0$ for points close to $\displaystyle (x,y)$ by the definition of continuity.