This question shouldn't be entirely that hard but for some reason when I check with my peers, all of us have completely varying or different answers... Here is the question. (Scanned the question from my worksheet)

So for

**PART A**, I did this.

$\displaystyle y' = 4x^3 - 10x^2y\frac{dy}{dx} - 10xy^2 + 16y^3\frac{dy}{dx} = 0$

$\displaystyle -10x^2y\frac{dy}{dx} +16y^3\frac{dy}{dx} = -4^3 + 10xy^2$

$\displaystyle \frac{dy}{dx}(16y^3 - 10x^2y) = 10xy^2 - 4x^3$

$\displaystyle \frac{dy}{dx} = \frac{10xy^2 - 4x^3}{16y^3 - 10x^2y} = \frac{5xy^2 - 2x^3}{8y^3 - 5x^2y}$

Then for

**PART B**, I plugged in (2,1) into my dy/dx and I got the slope to be 1/2 and thus my tangent line equation for $\displaystyle \ell$ was $\displaystyle y = \frac{1}{2}x + 0$

Then for

**PART C**, I set the equation to $\displaystyle y(2) = 0$ for $\displaystyle x=2$ and I got that $\displaystyle y = 1, -1, 2, -2$. But I'm not sure what to do with this...

So this is the work I did so far and I'm not sure on how to figure out part C or D of the question... If anyone could correct or check my initial work to see that it's correct or help me with Part C and D, then I would be really grateful! Because with some peers I was checking my work with, they got different derivatives or for part B got slopes like -3/14 or -13/14.