If $y=f(x)$ is a function derivative implicitly for the locus $y^3+2xy^2+x=4$, how I find $f(1)$??

Question

Question: Suppose that $y = f(x)$ is a derivative function given implicitly from the equation $y^3+2xy^2+x=4$. Suppose that, also, $1\in Df$

a) Compute $f(1)$.

b) Determine the equation of the tangent line to $f$ in $x=1$.

I don't know How to find $f(1)$. I know that $F(x,y)= 0$ defines implicitly a function derivative $y= f(x)$. But when I differentiate I have this:

\begin{gather} \frac{d}{dx}[y^3+2xy^2+x] = \frac{d}{dx}[4] \\ 3y^2\frac{dy}{dx}+2y^2+4xy\frac{dy}{dx}+1 =0 \\ 3y^2\frac{dy}{dx}+4xy\frac{dy}{dx} =-1-2y^2 \\ \frac{dy}{dx}=\frac{-1-2y^2}{3y^2+4xy} \end{gather}

Where do I deduce $f(1)$ from?

Because I need to determine the equation of the line, but I can't without this result.

Answer 1

You can find $f(1)$ from the original equation because it means finding the $y$-value when $x=1$.

$$y^3+2xy^2+x=4$$

Substitute in $x=1$ and you get

$$\begin{split} 0 &= y^3+2y^2-3\\ &=(y-1)(y^2+3y+3) \end{split}$$

Note that $\sqrt{3^2-4(3)}$ is complex so $y=1$ is the only choice.