\( \begin{array}{l}f(3)=2 f^{\prime}(3)=4 \\ \text { and } \\ g^{\prime}(3)=-3 g(3)=9\end{array} \rightarrow\left(\frac{f}{g}-f\right)^{\prime}(3)= \) ?

To find the derivative of \( \left(\frac{f}{g} - f\right) \) at \( x=3 \), we can start by applying the quotient rule and then simplifying. The first part is the derivative of the quotient: \[ \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \] Now substituting the known values at \( x=3 \): 1. \( f(3) = 2 \) 2. \( g(3) = 9 \) 3. \( f'(3) = 4 \) 4. \( g'(3) = -3 \) Plug these values into the derivative: \[ \left( \frac{f}{g} \right)'(3) = \frac{(4)(9) - (2)(-3)}{9^2} = \frac{36 + 6}{81} = \frac{42}{81} = \frac{14}{27} \] Now we find the derivative of \( -f \): \[ (-f)'(3) = -f'(3) = -4 \] Now combine these results to find \( \left( \frac{f}{g} - f \right)'(3) \): \[ \left( \frac{f}{g} - f \right)'(3) = \left( \frac{f}{g} \right)'(3) - (-f)'(3) = \frac{14}{27} - 4 \] To subtract, express 4 with a common denominator: \[ 4 = \frac{108}{27} \] Thus, we have: \[ \frac{14}{27} - \frac{108}{27} = \frac{14 - 108}{27} = \frac{-94}{27} \] Therefore, the final answer is: \[ \left(\frac{f}{g} - f\right)^{\prime}(3) = \frac{-94}{27} \]