The function $f$ is given by $f(x)=\frac{1}{3} e^{-3 x}+1$. What is the equation of the line tangent to the graph of $f$ when $x=-\ln 2 ?$

Step 1 ：\(f(x) = \frac{1}{3} e^{-3x} + 1\)

Step 2 ：\(f' (x) = \frac{d}{dx} \left( \frac{1}{3} e^{-3x} + 1 \right)\)

Step 3 ：\(f' (x) = -e^{-3x}\)

Step 4 ：\(x = -\ln 2\)

Step 5 ：\(f(-\ln 2) = \frac{1}{3} e^{-3(-\ln 2)} + 1\)

Step 6 ：\(f(-\ln 2) = \frac{1}{3} e^{3 \ln 2} + 1\)

Step 7 ：\(f(-\ln 2) = \frac{1}{3} (2^3) + 1\)

Step 8 ：\(f(-\ln 2) = \frac{8}{3} + 1\)

Step 9 ：\(f(-\ln 2) = \frac{11}{3}\)

Step 10 ：\(f' (-\ln 2) = -e^{-3(-\ln 2)}\)

Step 11 ：\(f' (-\ln 2) = -e^{3 \ln 2}\)

Step 12 ：\(f' (-\ln 2) = -2^3\)

Step 13 ：\(f' (-\ln 2) = -8\)

Step 14 ：\(y - \frac{11}{3} = -8(x + \ln 2)\)

Step 15 ：\boxed{y = -8x - 8 \ln 2 + \frac{11}{3}}