Let $g(n)$ denote the indicator function for the fourth powers. Then your sum equals
$$\sum_{n\leq x}\left(h_{r}*g\right)(n),$$
where $*$ denotes Dirichlet convolution. We may write rewrite the given asymptotic as
$$\sum_{k\leq x}g(k)\sum_{n\leq\frac{x}{k}}h_{r}(n)=2^{-1}\zeta(2)\zeta(4)x+O\left(x^{2/3+\epsilon}\right),$$
noticing that the right hand side this is of the form
$$G(x)=\sum_{n\leq x}\alpha(n)F\left(\frac{x}{n}\right).$$
Mobius inversion tells us that
$$F(x)=\sum_{n\leq x}\alpha^{-1}(n)G\left(\frac{x}{n}\right),$$
where $\alpha^{-1}$ is the multiplicative inverse of $\alpha$ with respect to Dirichlet convolution. Applying Mobius inversion to our sum, we have that
$$\sum_{n\leq x}h_{r}(n)=\sum_{j^{4}\leq x}\mu(j)\sum_{n\leq\frac{x}{j^{4}}}\left(h_{r}*g\right)(n)$$
which equals
$$2^{-1}\zeta(2)\zeta(4)x\sum_{j^{4}\leq x}\frac{\mu(j)}{j^{4}}+O\left(x^{2/3+\epsilon}\left(\sum_{j^{4}\leq x}\frac{1}{j^{4}}\right)\right)$$
$$=2^{-1}\zeta(2)x+O\left(x^{2/3+\epsilon}\right).$$
Notice that the Dirichlet inverse to the function $g(n)$, the indicator function for the fourth powers, is in some sense the mobius function on fourth powers.
