Let C be the curve `y^(3) - 3xy + 2 =0`. If H is the set of points on the curve C, where the tangent is horizontal and V is the set of points on the curve C, where the tangent is vertical, then H = … and V = … .

Answer

To solve the problem, we need to find the sets \( H \) and \( V \) for the curve defined by the equation \( y^3 - 3xy + 2 = 0 \).

### Step 1: Differentiate the curve equation
We start by differentiating the given equation with respect to \( x \):
\[ \frac{d}{dx}(y^3 - 3xy + 2) = 0 \]
Using the chain rule and product rule, we get:
\[ 3y^2 \frac{dy}{dx} - 3\left( y + x \frac{dy}{dx} \right) = 0 \]
This simplifies to:
\[ 3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0 \]

### Step 2: Rearranging the equation
We can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ (3y^2 - 3x) \frac{dy}{dx} = 3y \]
Thus, we find:
\[ \frac{dy}{dx} = \frac{3y}{3y^2 - 3x} = \frac{y}{y^2 - x} \]

### Step 3: Finding horizontal tangents (Set \( H \))
For the tangent to be horizontal, we set the derivative equal to zero:
\[ \frac{dy}{dx} = 0 \implies y = 0 \]
Now, we substitute \( y = 0 \) back into the original curve equation:
\[ 0^3 - 3x(0) + 2 = 0 \implies 2 = 0 \]
This is a contradiction, meaning there are no points on the curve where the tangent is horizontal. Thus, the set \( H \) is empty:
\[ H = \emptyset \]

### Step 4: Finding vertical tangents (Set \( V \))
For the tangent to be vertical, we set the denominator of the derivative equal to zero:
\[ y^2 - x = 0 \implies y^2 = x \implies y = \pm \sqrt{x} \]
Now, we substitute \( y = \sqrt{x} \) into the original curve equation:
\[ (\sqrt{x})^3 - 3x(\sqrt{x}) + 2 = 0 \implies x^{3/2} - 3x^{3/2} + 2 = 0 \implies -2x^{3/2} + 2 = 0 \]
This simplifies to:
\[ x^{3/2} = 1 \implies x = 1 \]
Now substituting \( x = 1 \) back to find \( y \):
\[ y = \sqrt{1} = 1 \quad \text{and} \quad y = -\sqrt{1} = -1 \]
Thus, we have two points: \( (1, 1) \) and \( (1, -1) \).

Next, we substitute \( y = -\sqrt{x} \) into the original curve equation:
\[ (-\sqrt{x})^3 - 3x(-\sqrt{x}) + 2 = 0 \implies -x^{3/2} + 3x^{3/2} + 2 = 0 \implies 2x^{3/2} + 2 = 0 \]
This also leads to:
\[ x^{3/2} = -1 \]
which has no real solutions.

Thus, the set \( V \) consists of the points where the tangent is vertical:
\[ V = \{(1, 1), (1, -1)\} \]

### Final Answer
- \( H = \emptyset \)
- \( V = \{(1, 1), (1, -1)\} \)