Given the function $f:\mathbb{R}^2\to\mathbb{R}$ defined by $$f(x,y)=x^3 + 3x^2y-y^3$$ find the points $(a,b)$ of the plane that satisfy the tangent of the level curve $M=f(a,b)$ in the point $(a,b)$ passes through $(0,1)$.
I tried solving this simply by using the equation of the tangent to a level curve: $$f_x(a,b)(x-a)+f_y(a,b)(y-b)=0$$
Hence, I get $$(3a^2+6ab)(x-a)+(3a^2-3b^2)(y-b) = 0$$
Then should I substitute $x,y$ for $(0,1)$? If so, after working it out, I get $$-a^3 -3a^2b+a^2-b^2+b^3=0$$ But how can I solve the points from here on out?
You're doing exactly the right thing. So this tells you that a point whose tangent contains $(0,1)$ must satisfy that last equation. It must also be a point of the original curve, i.e., satisfy $x^3 + 3x^2y - y^3 = M$. Those two equations together may determine just a few points.