Explanation

(b)
Given, $\frac{dy}{dx} = \frac{x^{3} + y^{3}}{xy^{2}}$
Put $y = vx$
$\Rightarrow$ $\frac{dy}{dx} = v + x\ \frac{dv}{dx}$
$\Rightarrow$ $x\frac{dv}{dx} + v = \frac{1 + v^{3}}{v^{2}}$
$\Rightarrow$ $v^{2}dv = \frac{dx}{x}$
$\Rightarrow$ $\frac{v^{3}}{3} = \log x + \log c$
$\Rightarrow$ $\frac{1}{3}\ \left( \frac{y}{x} \right)^{3} = \log x + \log c$
$\Rightarrow$ $y^{3} = 3x^{2}\log{cx}$