Implicit differentiation of $x^2+y^2-1$ $f(x,y) = x^2+y^2-1$
$0=x^2+y^2-1 \Rightarrow y=\sqrt{1-x^2}$
I differentiaded $g(x)=\sqrt{1-x^2}$ with the chain rule and got $g'(x)=-\frac{x}{\sqrt{1-x^2}}$.

Can someone tell me how to  do it with implicit differentiation?
I tried this formula $y'(x)=-\frac{f_x}{f_y}=-\frac{x}{y}$, the solution should obviously be the same so I guess I might not be allowed to use the formula here?
We had the implicit function theorem and
$dF(x,y)\begin{pmatrix}h\\k\\\end{pmatrix}=\begin{pmatrix}h\\df(x,y)(h,k)\\\end{pmatrix}$ but I don't really know how to apply this here.
 A: Notice that you have made the definition $g(x)=y$. Your implicit differentiation formula tells you that
$$y'=-\frac{x}{y}$$
but you said earlier
$$y=g(x)=\sqrt{1-x^2}$$
substituting this into your result for $y'$, you are left with
$$y'=-\frac{x}{\sqrt{1-x^2}}$$
so your method was correct, the "problem" is that you are left with a $y$ in your solution which is actually fine. If you dislike it, you can always solve for $y$ in the equation
$$0=x^2+y^2-1$$
and substitute it in, which we have already done.
A: All of the previous is correct.  However, the simpler way to do implicit differentiation is to just do it directly.  That is, use the differential instead of the derivative.  The differential is just the derivative without dividing by $dx$ at the end (i.e., there is no variable to take the derivative with respect to).  So, for instance, the differential $x^2$ is $2x\,dx$.  The differential of $w^2$ is $2w\,dw$.  So, applying that to your equation, we see:
$$ 0 = x^2 + y^2 - 1 \\ 
d(0) = d(x^2 + y^2 - 1) \\
d(0) = d(x^2) + d(y^2) - d(1) \\
0 = 2x\,dx + 2y\,dy - 0 \\
-2x\,dx = 2y\,dy \\
\frac{-x}{y} = \frac{dy}{dx}
$$
The second line simply applies the differential to both sides of the equation.  The third line uses the addition rule to break apart the differential at the plus boundaries.  The fourth line applies the differential.  The fifth and sixth lines simplify and solve for the ratio you are looking for ($\frac{dy}{dx}$).
