Answer
$${x^3}\ln \left| x \right| + C{x^3} = {y^3}$$
Work Step by Step
$$\eqalign{
& \left( {{x^3} + {y^3}} \right)dx - x{y^2}dy = 0 \cr
& {\text{Let }}y = ux,{\text{ }}dy = udx + xdu \cr
& {\text{Substituting}} \cr
& \left( {{x^3} + {{\left( {ux} \right)}^3}} \right)dx - x{\left( {ux} \right)^2}\left( {udx + xdu} \right) = 0 \cr
& \left( {{x^3} + {u^3}{x^3}} \right)dx - {u^2}{x^3}\left( {udx + xdu} \right) = 0 \cr
& \left( {1 + {u^3}} \right){x^3}dx - {u^3}{x^3}dx - {u^2}{x^4}du = 0 \cr
& \left[ {\left( {1 + {u^3}} \right) - {u^3}} \right]{x^3}dx = {u^2}{x^4}du \cr
& {x^3}dx = {u^2}{x^4}du \cr
& {\text{Separate the variables}} \cr
& \frac{{{x^3}}}{{{x^4}}}dx = {u^2}du \cr
& \frac{1}{x}dx = {u^2}du \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{1}{x}} dx = \int {{u^2}} du \cr
& \ln \left| x \right| + C = \frac{1}{3}{u^3} \cr
& y = ux \to u = \frac{y}{x} \cr
& \ln \left| x \right| + C = \frac{1}{3}{\left( {\frac{y}{x}} \right)^3} \cr
& \ln \left| x \right| + C = \frac{{{y^3}}}{{{x^3}}} \cr
& {x^3}\ln \left| x \right| + C{x^3} = {y^3} \cr} $$
