For the function $y=f(x)=5 x^{3}+4$ : a. Find $\frac{d f}{d x}$ at $x=-9$. \[ f^{\prime}(-9)= \] b. Find a formula for $x=f^{-1}(y)$. \[ f^{-1}(y)= \] c. Find $\frac{d f^{-1}}{d y}$ at $y=f(-9)$ \[ \left(f^{-1}\right)^{\prime}(f(-9))= \]

Step 1 ：Find the derivative of the function \(f(x) = 5x^3 + 4\). The derivative of a constant is zero and the derivative of \(x^n\) is \(nx^{n-1}\). So, the derivative of \(f(x)\) is \(f'(x) = 15x^2\).

Step 2 ：Evaluate \(f'(x)\) at \(x=-9\).

Step 3 ：Find the inverse function \(f^{-1}(y)\). To do this, solve the equation \(y = 5x^3 + 4\) for \(x\).

Step 4 ：Find the derivative of the inverse function at \(y=f(-9)\). The formula for the derivative of the inverse function is \((f^{-1})'(y) = 1/(f'(f^{-1}(y)))\).

Step 5 ：The derivative of the function \(f(x) = 5x^3 + 4\) at \(x=-9\) is \(\boxed{1215}\).

Step 6 ：The inverse function \(f^{-1}(y)\) is \(\boxed{\left(\frac{y}{5} - \frac{4}{5}\right)^{\frac{1}{3}}}, \boxed{-\frac{1}{2}\left(\frac{y}{5} - \frac{4}{5}\right)^{\frac{1}{3}} - \frac{\sqrt{3}i}{2}\left(\frac{y}{5} - \frac{4}{5}\right)^{\frac{1}{3}}}, \boxed{-\frac{1}{2}\left(\frac{y}{5} - \frac{4}{5}\right)^{\frac{1}{3}} + \frac{\sqrt{3}i}{2}\left(\frac{y}{5} - \frac{4}{5}\right)^{\frac{1}{3}}}\).

Step 7 ：The derivative of the inverse function at \(y=f(-9)\) is \(\boxed{\frac{1}{15}\left(\frac{y}{5} - \frac{4}{5}\right)^{-\frac{2}{3}}}\).