for the curve C x^{3} + xy + 2y^{3} =k , the y co-ordinates of the tangent parallel to the y axis is given by
216y^{6 }+4y^{3} +k = 0
Show that k<= 1/54
Let $\displaystyle m = y^3$, so the second equation becomes
$\displaystyle 216m^2 + 4m + k = 0$
The discriminant of this quadratic is $\displaystyle 16 - 4(216)(k) = 16 - 864k$
For the quadratic to even have real solutions for m, the discriminant must be non-negative, i.e. $\displaystyle 16 - 864k \ge 0 \Rightarrow k \le \frac{1}{54}$. No calculus needed.
For the other question, differentiate both sides of the original equation:
$\displaystyle 3x^2 + (y + x\frac{dy}{dx}) + 6y^2 \frac{dy}{dx} = 0$
$\displaystyle \frac{dy}{dx} (x + 6y^2) + 3x^2 + y = 0$
$\displaystyle \frac{dy}{dx} = -\frac{3x^2 + y}{x + 6y^2}$
If the line $\displaystyle x = -6$ is a tangent at the point (x,y), it means that x = -6 and dy/dx is undefined, i.e. the denominator of dy/dx is zero. This occurs only when $\displaystyle y = \pm 1$. So try substituting (1,6) and (-1,6) into your original equation and find k.