The question is
Suppose that $$\sum_{k=1}^\infty d_3(k)|f(kx)|<\infty,$$
where $d_3(k)$ denotes the number of factorizations of $k$ as a product of three numbers. Show that if $$g(x)=\sum_{m=1}^\infty f(mx),$$ then $$f(x)=\sum_{n=1}^\infty \mu(n)g(nx)$$ and conversely.
I don't see how to prove that $$\sum_{m,n} g(mnx)$$ converges absolutely, which we need to show the converse statement in the exercise. The solution in the book says it follows from the convergence of $$\sum_{k=1}^\infty d_3(k)|f(kx)|<\infty,$$ but I don't see how to do this without assuming $$g(x)=\sum_{m=1}^\infty f(mx),$$ which would be circular.
Don't worry, your reasoning is correct. The converse is definitely false.
Let $g(x)=\frac{1}{x},$ so that $$\sum_{k=1}^{\infty}\mu(k)g(kx)=\frac{1}{x}\sum_{k=1}^{\infty}\frac{\mu(k)}{k}=0.$$ Then $f(x)=0,$ and hence the sum $\sum_{k=1}^{\infty}d_{3}(k)|f(kx)|$ converges, and equals $0$. However, we obviously don't have $$\sum_{k=1}^{\infty}f(kx)=g(x).$$ To have the converse, we must assume that $$\sum_{k=1}^{\infty}d_{2}(k)|g(kx)|<\infty,$$ and the condition $\sum_{k=1}^{\infty}d_{3}(k)|f(kx)|$ is only sufficient in the forward direction. In the case above, the series did in fact diverge since $$\sum_{k=1}^{\infty}d_{2}(k)|g(kx)|=\frac{1}{x}\sum_{k=1}^{\infty}\frac{d_{2}(k)}{k}\gg\sum_{k=1}^{\infty}\frac{1}{k}.$$