find dy/dx by implicit differentiation: x^2y^3-7xy^2=10

[tex]x^2y^3-7xy^2=10[/tex]Differentiate both sides wrt [tex]x[/tex]:[tex]\dfrac{\mathrm d(x^2y^3-7xy^2)}{\mathrm dx}=\dfrac{\mathrm d(10)}{\mathrm dx}[/tex][tex]\dfrac{\mathrm d(x^2y^3)}{\mathrm dx}-7\dfrac{\mathrm d(xy^2)}{\mathrm dx}=0[/tex]By the product rule,[tex]\left(y^3\dfrac{\mathrm d(x^2)}{\mathrm dx}+x^2\dfrac{\mathrm d(y^3)}{\mathrm dx}\right)-7\left(y^2\dfrac{\mathrm d(x)}{\mathrm dx}+x\dfrac{\mathrm d(y^2)}{\mathrm dx}\right)=0[/tex]By the power and chain rules,[tex]\left(y^3(2x)+x^2\left(3y^2\dfrac{\mathrm dy}{\mathrm dx}\right)\right)-7\left(y^2+x\left(2y\dfrac{\mathrm dy}{\mathrm dx}\right)\right)=0[/tex][tex]\left(2xy^3+3x^2y^2\dfrac{\mathrm dy}{\mathrm dx}\right)-7\left(y^2+2xy\dfrac{\mathrm dy}{\mathrm dx}\right)=0[/tex][tex](3x^2y^2-14xy)\dfrac{\mathrm dy}{\mathrm dx}=7y^2-2xy^3[/tex][tex]\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{7y^2-2xy^3}{3x^2y^2-14xy}[/tex]and if [tex]y\neq0[/tex],[tex]\implies\boxed{\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{7y-2xy^2}{3x^2y-14x}}[/tex]