Answer
$$g(x(t), y(t))=\frac{\sqrt{t}}{\left(t^{2}+1\right)^{3}}$$
Work Step by Step
Given:
\[
\begin{array}{c}
y e^{-3 x}=g(x, y) \\
\ln \left(1+t^{2}\right), y(t)=\sqrt{t}=x(t)
\end{array}
\]
We find:
\[
\because y e^{-3 x}=g(x, y)
\]
\[
\begin{aligned}
g(x(t), y(t)) &=y(t) e^{-3 x(t)} \\
g(x(t), y(t)&=\sqrt{t} * e^{-3 \ln \left(t^{2}+1\right)} \\
g(x(t), y(t))&=\frac{\sqrt{t}}{\left(t^{2}+1\right)^{3}}
\end{aligned}
\]
