[tex]\bf 2x^2+xy+y^2=36\\\\
-----------------------------\\\\
4x+\left[1\cdot y+x\frac{dy}{dx} \right]+2y\frac{dy}{dx}=0
\\\\\\
\cfrac{dy}{dx}(x+2y)=-4x-y\implies \left. \cfrac{dy}{dx}=\cfrac{-4x-y}{x+2y} \right|_{(3,3)}\implies -\cfrac{5}{3}\leftarrow m\\\\
-----------------------------\\\\
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-3=-\cfrac{5}{3}(x-3)\\
\qquad \uparrow\\
\textit{point-slope form}[/tex]
and solve that for "y", that'd be the equation for the tangent line at 3,3
