Derivative of a lemniscate at the left hand side How does $${d \over{dx}}(3(x^2+y^2)^2)$$ turn into $$12y(x^2+y^2){{dy} \over{dx}}+12x(x^2+y^2)$$? I'm having a hard time solving it algebraically without it turning into a huge polynomial.

Use the chain rule.
$$3\left({d\over{dx}}(x^2+y^2)^2\right)=3\cdot2{d\over{dx}}(x^2+y^2)(x^2+y^2)=6\left(2x+2{d\over{dx}}(y)y\right)(x^2+y^2)=6\left(2x+2y{dy\over{dx}}\right)(x^2+y^2)=12(x^2+y^2)\left(y{dy\over{dx}}+x\right)$$
