Step 1: See the structure.
We need $\int x^5e^{-4x^3}\,dx$ written as $\tfrac{1}{48}e^{-4x^3}f(x)+c$, and our job is to find $f(x)$. The power $x^3$ inside the exponential suggests a substitution.
Step 2: Substitute.
Let $t=x^3$, so $dt=3x^2\,dx$. Split $x^5=x^3\cdot x^2$. Then \[ \int x^3e^{-4x^3}\cdot x^2\,dx=\frac{1}{3}\int t\,e^{-4t}\,dt. \]
Step 3: Set up integration by parts.
For $\int t\,e^{-4t}\,dt$, take $u=t$ and $dv=e^{-4t}\,dt$. Then $du=dt$ and $v=-\tfrac{1}{4}e^{-4t}$.
Step 4: Apply the parts formula.
\[ \int t\,e^{-4t}\,dt=-\frac{t}{4}e^{-4t}+\frac{1}{4}\int e^{-4t}\,dt=-\frac{t}{4}e^{-4t}-\frac{1}{16}e^{-4t}. \]
Step 5: Multiply by the outside $\tfrac{1}{3}$.
\[ I=\frac{1}{3}\left(-\frac{t}{4}-\frac{1}{16}\right)e^{-4t}=\frac{1}{48}e^{-4t}(-4t-1). \]
Step 6: Put $t=x^3$ back.
So $I=\tfrac{1}{48}e^{-4x^3}(-4x^3-1)$. Comparing with the given form, \[ f(x)=-4x^3-1. \]
\[ \boxed{-4x^3-1} \]