\((6x+1)y^2 dy/dx+3x^2+2y^3=0\)

\((6x+1)y^2dy+(3x^2+2y^3)dx=0\)

\((6xy^2+1y^2)dy+(3x^2+2y^3)dx=0\)

Comparing with \(mdx+ndy=0\)

\(n= (6xy^2+1y^2)\)

\(m= (3x^2+2y^3)\)

\((del\ m)/(del\ y)=6y^2\)

\((del\ n)/(del\ x)=6y^2\)

\(\int\ mdx+ \int\) terms of n not involving x dy= C(x)

\(\int (3x^2+2y^3)dx+ \int y^2 dy= C(x)\)

\(x^3+2xy^3+y^3/3= C(x)\)

Now applying initial conditions

Substituting the value for C

y(0)=1

\(x^3+2xy^3+y^3/3= C(x)\)

\(C=1/3\)

\(x^3+2xy^3+y^3/3=1/3\)