Answer is: option4

\( y = \frac{7}{8}x + \frac{9}{8} \)

Solution:

The equation of the curve is \( 3y^2 - x^3 - xy^2 = 7 \).

1. Differentiate implicitly with respect to \( x \):

\[ 6y \frac{dy}{dx} - 3x^2 - y^2 - 2xy \frac{dy}{dx} = 0 \]

2. Factor \( \frac{dy}{dx} \) terms:

\[ \frac{dy}{dx}(6y - 2xy) = 3x^2 + y^2 \]

3. Solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{3x^2 + y^2}{6y - 2xy} \]

4. Substitute the point \( (1, 2) \) into the equation:

\[ \frac{dy}{dx} \text{ at } (1,2) = \frac{3(1)^2 + 2^2}{6(2) - 2(1)(2)} = \frac{3 + 4}{12 - 4} = \frac{7}{8} \]

5. The slope of the tangent is \( \frac{7}{8} \). The equation of the tangent line at \( (1, 2) \) is:

\[ y - 2 = \frac{7}{8}(x - 1) \]

6. Expanding this equation:

\[ y = \frac{7}{8}x + 2 - \frac{7}{8} = \frac{7}{8}x + \frac{9}{8} \]

Therefore, the equation of the tangent line is:

\[ y = \frac{7}{8}x + \frac{9}{8} \]