Find equation of tangent line using differential equation: dy/dx = x(y^1/3) 
The expression $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x} = x\sqrt[3]{y}$ gives the slope at any point on the graph of the function $f(x)$ where $f(2) = 8$.
a. Write the equation of the tangent line to $f(x)$ at point $(2,8)$.

Since $\frac{\mathrm{d}y}{\mathrm{d}x}$ gives the slope of the tangent line at any point, can I just plug and chug $(2,8)$ into the differential equation, and then use that slope in the point-slope formula? This would give:
$\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x} = (2)(\sqrt[3]{8})$
$\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x} = 4$
$y - 8 = 4(x - 2)\\
y = 4(x - 2) + 8$ 
... could anyone verify that?

b. Write an expression for $f(x)$ in terms of $x$.

I got: $x = \sqrt{3y^\frac 23 + C}$, where $C$ is a constant of integration. Not quite sure whether this is correct or not.

c. What is the domain of $f(x)$?

I have no idea how to begin answering this question. Could it be all real numbers, since the domain of the differential equation seems to be all real numbers?

d. What is the minimum value of $f(x)$?

I think I have to set $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x} = 1$, but then what? Is it $x = 0, y = 0$?

separating the variables and integrating, we get $$\int_8^y y^{-1/3}\, dy = \int_2^x x \,dx \to \frac32\left(y^{2/3} - 4\right)=\frac12\left(x^2 - 4\right) $$ solving for $$y = f(x) = \left(\frac 13 (x^2+8)\right)^{3/2}.$$ that answers part(b).
for part (c), the domain of $f$ is all real numbers.
and for part (d), $f$ has a global minimum $f(0) = \left(\frac 83\right)^{3/2}$ at $x = 0.$
