Interval of validity of an ODE 
Please help me with parts (c) and (d). 
Part (c) says that we need to find the interval of validity, do I need to put x=0 and then check for values of y? I already did that and got y = 0, +2, -2. So, that means that the interval of validity is [0,0], [-2,0], and [0,2] but I can see that the ODE has a vertical tangent at y= +2/sqrt(3), and -2/sqrt(3). That basically means that the above intervals I mentioned is not valid. 
So, what should I do??
Also, for part (d), I dont have any idea what to do. Please help!!
 A: Imagine for a moment the related equation
$$x'(y) = \frac{3y^2-4}{3x^2}$$
At the place where the previous function had a vertical tangent, this one has a critical point. Let's take the second derivative to find out what kind of critical point it is:
$$x''(y) = \frac{3x^2(6y)-(3y^2-4)6xx'}{9x^4} = \pm\frac{2|y|}{x^2} \neq 0$$
at the critical points. The exact value is not important, only the fact that it is not $0$. A differentiable function with a local min or a max is not invertible since the inverse would fail the vertical line test. 
This means that the specific curve we want that contains $(0,0)$ only goes up to the $x$ value that gives the vertical tangent. Solving:
$$x^3 = y(y^2-4) \implies x = \mp \frac{2\sqrt[3]{2}}{\sqrt{3}}$$
and your interval of convergence is on
$$\left[-\frac{2\sqrt[3]{2}}{\sqrt{3}}, \frac{2\sqrt[3]{2}}{\sqrt{3}}\right]$$
which contains $0$. The instructor had you draw a slope field to avoid this chain of reasoning because it only makes rigorous what the drawing makes obvious.
