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Solution:

First, we show that $\lim_{x\to 0^+}f(x)$ exists if $\lim_{y\to \infty}g(y)$ exists.

Suppose $\lim_{y\to \infty}g(y)=L$. For any $\epsilon>0$, there exists $N>0$ such that
\begin{equation}\label{30-4-1}
|g(y)-L|<\epsilon,
\end{equation} for any $y>N$.

Let $\delta=\dfrac{1}{N}$.

Note that for all $0 < x < \delta$, we have $\dfrac{1}{x} > N$. Therefore, by \eqref{30-4-1},
$$
|f(x)-L|=\left|g\Big(\frac{1}{x}\Big)-L\right|<\epsilon
$$ for all $0 < x < \delta$. Therefore, we have $\lim_{x\to 0^+}f(x)=L$.

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Then, we show the other direction. Namely $\lim_{x\to 0^+}f(x)$ exists only if $\lim_{y\to \infty}g(y)$ exists.

Suppose $\lim_{x\to 0^+}f(x)=L$. For any $\epsilon >0$, there exists $\delta>0$ such that
\begin{equation}\label{30-4-2}
|f(x)-L|<\epsilon,
\end{equation} for all $0< x< \delta$.

Let $N=\max\{\dfrac{1}{\delta},a^{-1}\}$.

Note that for all $y>N$, we have $0<\dfrac{1}{y}<\delta$. Therefore, by \eqref{30-4-2}, we have
$$
|g(y)-L|=\left|f\Big(\frac{1}{y}\Big)-L\right|<\epsilon
$$ for all $y>N$. Therefore, we have $\lim_{y\to \infty}g(y)=L$.