= Solution
Put $y=\lambda x$ and $t=y+s$. Factoring out the endpoint value gives
$$
\gamma(x,y)=y^{x-1}e^{-y}
\int_0^\infty
\exp\left((x-1)\log\left(1+\frac{s}{\lambda x}\right)-s\right)ds.
$$
For fixed $s$,
$$
(x-1)\log\left(1+\frac{s}{\lambda x}\right)-s
=-\frac{\lambda-1}{\lambda}s
-\frac1x\left(\frac{s}{\lambda}+\frac{s^2}{2\lambda^2}\right)
+O(x^{-2}).
$$
Watson's endpoint lemma justifies termwise integration, so with $\alpha=(\lambda-1)/\lambda$,
$$
\begin{aligned}
\frac{\gamma(x,y)}{y^{x-1}e^{-y}}
&\sim\int_0^\infty e^{-\alpha s}
\left[1-\frac1x\left(\frac{s}{\lambda}
+\frac{s^2}{2\lambda^2}\right)\right]ds\\
&=\frac1\alpha
-\frac1x\left(\frac1{\lambda\alpha^2}
+\frac1{\lambda^2\alpha^3}\right).
\end{aligned}
$$
Consequently
$$
\boxed{
f(\lambda)=\frac{\lambda}{\lambda-1},
\qquad
g(\lambda)=-\frac{\lambda^2}{(\lambda-1)^3}.}
$$
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