[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac16x(1-y^2)[/tex]
[tex]\displaystyle\int\frac{\mathrm dy}{1-y^2}=\frac16\int x\,\mathrm dx[/tex]
[tex]\dfrac12\ln|1+y|-\dfrac12\ln|1-y|=\dfrac1{12}x^2+C[/tex]
[tex]\ln\left|\dfrac{1+y}{1-y}\right|=\dfrac16x^2+C[/tex]
[tex]\dfrac{1+y}{1-y}=e^{x^2/6+C}[/tex]
[tex]-1+\dfrac2{1-y}=Ce^{x^2/6}[/tex]
[tex]y=1-\dfrac2{1+Ce^{x^2/6}}[/tex]
Given that [tex]y(0)=7[/tex], we get
[tex]7=1-\dfrac2{1+Ce^0}\implies C=-\dfrac43[/tex]
so the particular solution is
[tex]y=1-\dfrac2{1-\frac43e^{x^2/6}}[/tex]
