Show that the equation $x^3 - 2x + 2y^3 + y^2 = 0$ defines a curve of the form ${(x, y) : x = \varphi(y)}$ in $B^2 ((0, 0), \delta)$ form some $\delta > 0$. Find the equation of the tangent at $(0,0)$
The first part is just an application of the implicit function theorem.
Since $F(x,y) = x^3 - 2x + 2y^3 + y^2$ and $F(0,0) = 0$, $\frac{\partial F}{\partial x}(0,0) = -2 \ne 0.$ There exist an $x = \varphi(y)$.
For the second part I'm familiar that the derivative for $\varphi(y)$ is $\varphi'(y) = - \frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial x}}$, but I'm not sure if this is the tangent they're referring to?