Answer is: option4

\( \frac{3}{2} \)

Solution:

We are given that the equation of the normal line to the graph of \( f \) at the point \( (-1,2) \) is:

\[ 2x + 3y = 4 \]

We need to determine \( f'(-1) \), which is the slope of the tangent line at \( x = -1 \).

Rearrange the equation \( 2x + 3y = 4 \) into slope-intercept form (\( y = mx + b \)):

\[ 3y = -2x + 4 \] \[ y = -\frac{2}{3}x + \frac{4}{3} \]

Thus, the slope of the normal line is:

\[ m_{\text{normal}} = -\frac{2}{3} \]

The tangent line is perpendicular to the normal line. The slopes of perpendicular lines satisfy:

\[ m_{\text{tangent}} = -\frac{1}{m_{\text{normal}}} \]

Substituting \( m_{\text{normal}} = -\frac{2}{3} \):

\[ m_{\text{tangent}} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} \]

Since \( m_{\text{tangent}} = f'(-1) \), we conclude:

\[ \boxed{\frac{3}{2}} \]

Thus, the correct answer is \( \frac{3}{2} \).