Finding equations of two lines through $(0,0)$ tangent to $x^2-4x+y^2+3=0$ I need to find equations of two lines that go through $(0,0)$ and tangent to the graph $$x^2-4x+y^2+3 = 0$$
So 
$$2x - 4 +2y\cdot y^\prime = 0 \qquad\rightarrow\qquad y^\prime = \frac{4-2x}{2y} $$
The equation of a line should be $y - y_0 = m(x-x_0)$, and $(0,0)$ should be a point on that line. So $y - 0 = m(x-0)$, where $m$ is the slope of the line, which I should just place the $y^\prime$ instead. Then I get 
$$y -0 = \frac{2-x}{y} (x-0) \qquad\rightarrow\qquad y = \frac{x(2-x)}{y}$$ 
But it is not the answer, as the answer is $$y = \pm \frac{3}{\sqrt{x}}$$
 A: 
(...) then I get $\ y -0 = \frac{2-x}{y} (x-0) \rightarrow y = \frac{x(2-x)}{y} $
but it is not the answer as the answer (...)

It can't be the answer because the expression you write is not the equation of a line! The slope is a number so you still need to evaluate the expression you obtained for $y'$ via implicit differentiation. Where? Well precisely in the point(s) where it is tangent to the given equation, but you don't know where that is (yet)!

Alternatively: the equation of a line passing through the origin $(0,0)$ is of the form $y=mx$. Substituting into $x^2-4x+y^2+3 = 0$, you get a quadratic equation in $x$ with parameter $m$:
$$x^2-4x+(mx)^2+3 = 0 \iff \left(1+m^2\right)x-4x+3=0$$
The line "touches" the graph (i.e. it is tangent) if this equation has exactly one solution, so you want the discriminant $D=b^2-4ac$ to be $0$ which easily gets you the values for $m$, without the need for differentiation:
$$\underbrace{16-12(1+m^2)}_{D=b^2-4ac}=0 \iff  m = \pm\frac{1}{\sqrt{3}}$$
This is probably what you meant instead of:

as the answer is $\ y = \pm  \frac{3}{\sqrt{x}} $

which isn't the equation of a line either.
Here's a plot to illustrate the result.

Addition after comment.
Look at the graph: an arbitrary line will usually either have no points of intersection with the circle (the corresponding equation in $x$ will then have $0$ solutions; strictly negative discriminant) or two points of intersection (the corresponding equation in $x$ will then have $2$ solutions; strictly positive discriminant).
In the (somewhat special) case of a tangent line, there is exactly one point of intersection (the corresponding equation in $x$ will then have $1$ solution; zero discriminant). Expressing this condition (on the discriminant of the quadratic equation in $x$) gives you two possible solutions for the parameter $m$ (not $x$!).
So you have two (different) cases with slopes $m$, corresponding to lines with exactly one point of intersection with the circle; so don't mix:

*

*you require the quadratic equation in $x$ to have one solution by setting the discriminant $0$;

*the equation for the discriminant being $0$ then yields two possible values for the slope $m$.

