$f^{\text{'}}(-2)=11\mathrm{.}245=m$

Q$.$ The graph of a bifolium is given as $(x^2+y^2{)}^2=4x^{2}y.$

Find the equation of the tangent line at the point $(-1,1).$

$u=x^2+y^2$

$dv=2xdx+2ydy$

Got stuct here, also we need to use $U$ substitution and not chan rule,

$2$vdu$=4(x^2(1)+2x(y))$

$2(x^2+y^2)(2x+2y)=4x^2+5xy$

$(4x+2y^2)(2x+2y)$

$4x^3+4x^{2}y+4x{y}^2+4y^3=4x^2+8xy$

thanks