[tex]y\ln(x^2+y^2+8)=3[/tex]
Differentiate both sides with respect to [tex]x[/tex]:
[tex]\dfrac{\mathrm dy}{\mathrm dx}\ln(x^2+y^2+8)+y\dfrac{\frac{\mathrm d}{\mathrm dx}[x^2+y^2+8]}{x^2+y^2+8}=0[/tex]
We have by the chain rule,
[tex]\dfrac{\mathrm d}{\mathrm dx}[x^2+y^2+8]=2x+2y\dfrac{\mathrm dy}{\mathrm dx}[/tex]
so that
[tex]\dfrac{\mathrm dy}{\mathrm dx}\left(\ln(x^2+y^2+8)+\dfrac{2y^2}{x^2+y^2+8}\right)=-\dfrac{2xy}{x^2+y^2+8}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2xy}{(x^2+y^2+8)\left(\ln(x^2+y^2+8)+\frac{2y^2}{x^2+y^2+8}\right)}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2xy}{(x^2+y^2+8)\ln(x^2+y^2+8)+2y^2}[/tex]
