As OP already found, the expression for the slope of a tangent line to a point on the curve in question is
$$ y \ ' \ = \ \frac{y^2 \ - \ 3x^2}{3y^2 \ - \ 2xy} \ = \ \frac{y^2 \ - \ 3x^2}{y \ (3y \ - \ 2x)} \ \ . $$
Something useful to check for immediately is whether this rational function can be "indeterminate", that is, whether both numerator and denominator can be zero at the same time for a point on the curve. There are two cases to consider, due to the two factors in the denominator:
I -- $ \ y^2 \ = \ 3x^2 \ \ , \ \ y \ = \ 0 \ \ \Rightarrow \ \ x \ = \ 0 \ $ ; however, the point $ \ ( 0 , 0 ) \ $ does not lie on this curve
II -- $ \ y^2 \ = \ 3x^2 \ \ , \ \ 3y \ = \ 2x \ \ \Rightarrow \ \ \left( \frac{2}{3}x \right)^2 \ = \ 3 x^2 \ \ \Rightarrow \ \ x \ = \ 0 \ \ , \ \ y \ = \ 0 \ $ ; again, this is not a solution to the curve equation.
So the issue of "slope indeterminancy" does not occur for this curve, as it does for one of its "cousins", the folium of Descartes (discussed, for instance, here).
We can then locate the points at which horizontal tangents occur by inserting the relation $ \ y^2 \ = \ 3x^2 \ \Rightarrow \ y \ = \ \pm \ 3^{1/2}x \ $ into the equation for the curve (as mentioned by Graham Kemp), thus:
$$ x^3 \ + \ y^3 \ - \ xy^2 \ = \ 8 \ \ \Rightarrow \ \ x^3 \ + \ \left( \pm \ 3^{1/2}x \right)^3 \ - \ x \ \left( \pm \ 3^{1/2}x \right)^2 \ = \ 8 $$
$$ \Rightarrow \ \ \left( 1 \ \pm \ 3^{3/2} \ - \ 3 \right) \ x^3 \ = \ 8 $$
$$ \Rightarrow \ \ x \ = \ \left[ \frac{8}{-2 \ \pm \ 3 \sqrt{3}} \right]^{1/3} = \ \frac{2}{( -2 \ \pm \ 3 \sqrt{3} \ )^{1/3}} \ \approx \ -1.036 \ , \ 1.358 $$
$$ y \ = \ \pm \ \sqrt{3} \ \cdot \ x \ \rightarrow \ 1.794 \ , \ 2.352 \ \ . $$
[Upon insertion into the original curve equation, we see that the negative values for $ \ y \ $ are not solutions.] Thus, the horizontal tangent lines are found at
$$ \left( \frac{2}{( -2 \ - \ 3 \sqrt{3} \ )^{1/3}} \ , \ \frac{2 \ \sqrt{3}}{( -2 \ - \ 3 \sqrt{3} \ )^{1/3}} \right) \ \ , \ \ \left( \frac{2}{( -2 \ + \ 3 \sqrt{3} \ )^{1/3}} \ , \ \frac{2 \ \sqrt{3}}{( -2 \ + \ 3 \sqrt{3} \ )^{1/3}} \right) \ \ . $$
Vertical tangents occur at the points for which one of the factors in the denominator of the expression for $ \ y \ ' \ $ equals zero:
$$ y \ = \ 0 \ \ \Rightarrow \ \ x^3 \ + \ 0^3 \ - \ x \cdot 0^2 \ = \ 8 \ \ \Rightarrow \ \ x \ = \ 2 \ \ ; $$
$$ y \ = \ \frac{2}{3}x \ \ \Rightarrow \ \ x^3 \ + \ \left( \frac{2}{3}x \right)^3 \ - \ x \cdot \left( \frac{2}{3}x \right)^2 \ = \ 8 $$
$$ \Rightarrow \ \ \left( 1 \ + \ \frac{8}{27} \ - \ \frac{4}{9} \right) \ x^3 \ = \ \left( \frac{27 \ + \ 8 \ - 12}{27} \right) x^3 \ = \ 8 $$
$$ \Rightarrow \ \ x \ = \ \left( \frac{ 8 \ \cdot \ 27}{23} \right)^{1/3} \ = \ \frac{6}{23^{1/3}} \ \approx \ 2.110 \ \ \Rightarrow \ \ y \ = \ \frac{2}{3} \cdot \ \frac{6}{23^{1/3}} \ = \ \frac{4}{23^{1/3}} \ \approx \ 1.407 \ \ . $$
So the points where vertical tangents are found are
$$ ( \ 2, 0 \ ) \ \ \text{and} \ \ \left( \frac{6}{23^{1/3}} , \frac{4}{23^{1/3}} \right) \ \ . $$
A graph of the curve is presented below.
horizontal tangents are marked in green, vertical tangents, in red
For part (a) , a critical is any point where the derivative is 0 or undefined. Therefore, y = 0 is also a critical point. However, it doesn't exist because there is no value of x that satisfies that value of y.
$$-4x^2 =2 $$
$$x^2 = \frac{-1}{2}$$
Which cannot exist.
When x = 0
$$-y^2 = -2$$
$$y = \sqrt{2}, -\sqrt{2}$$
For part (b) then,
$$\frac{4y^2 - 16x^2}{y^3}$$
When $x = 2, y = \sqrt{2}$
$$\frac{4*2 - 0}{2\sqrt{2}}$$
$$\frac{4}{\sqrt{2}}$$
$$2\sqrt{2}$$
When $x = 2, y = -\sqrt{2}$
$$\frac{4*2 - 0}{-2\sqrt{2}}$$
$$\frac{4}{-\sqrt{2}}$$
$$-2\sqrt{2}$$
When drawing the graph, use values of x and y that yield integer values. From (a) and (b), you can guess the concavity at the critical points of the functions, or the point where the slope is 0. However, please note that if the derivative is undefined it yields a vertical tangent line, if it is 0 then it yields a horizontal tangent line or a slope of 0.
If you need more help, please ask as a comment.
Best Answer
You correctly found that \begin{equation} \frac{d y}{d x} = - \frac{x}{4y} \end{equation} using implicit differentiation. The definition of a stationary point $(x^*, y^*)$ is that it lies on the curve and that \begin{equation} \left.\frac{d y}{d x}\right|_{x = x^*, y = y^*} = 0. \end{equation} Thus, we want $-\frac{x}{4y} = 0$. This happens when $x = 0$ regardless of the value of $y$. We also have to make sure that the stationary point(s) lies on the curve. To this end, plug back into the equation for the curve to see that at the stationary point(s) we must have \begin{equation} \frac{y^2}{5} = 1. \end{equation} Therefore, the two stationary points of the curve are $(0, \sqrt{5})$ and $(0, -\sqrt{5})$.