Find x coordinates of the horizontal tangent line.

Question

This is the tangent line of a function. How can I find all the x coordinates of the line tangent to the original function?

$$y'={2x\over3y^2 +2y-5}$$

Answer 1

Notice, in general, $y'$ shows the slope of tangent to the curve at any point $(x, y)$. Since, the tangent line is horizontal hence it's parallel to the x-axis i.e. its slope is $0$ hence setting $y'=0$ in the given expression, one should get $$y'=\frac{2x}{3y^2+2y-5}=0$$
$$\frac{2x}{(y-1)(3y+5)}=0$$

$$\color{red}{x=0}$$$$\ \ \ \forall \ \ \ y\ne 1\ \ \ \text{&}\ \ \ y\ne -\frac{5}{3}$$

Answer 2

Multiply by the bottom:

$3y^2y'+2yy'-5y'=2x$

And, integrade.

$y^3+y^2-5y=x^2$

Then, you can plug in the needed $y$ value to get the value for $x^2$, and take square root to find the $x$ coordinate.