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Let me elaborate on Matt Young's comment and show, for a sufficiently small absolute constant $c>0$, $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} \ll \exp(-c\sqrt{\log z}).$$ All references will be from Montgomery-Vaughan: Multiplicative number theory I.
We can assume that $z>2$ is not an integer. The associated Dirichlet series $$F(s):=\sum_{d=1}^\infty\mu(d)\frac{\tau(d)}{d^{s+1}} =\prod_p\left(1-\frac{2}{p^{s+1}}\right)=\frac{G(s)}{\zeta(s+1)^2}$$ is absolutely convergent in $\Re s>0$, and $G(s)$ is given by an absolutely convergent Euler product in $\Re s>-1/2$. In particular, $G(s)$ is holomorphic, bounded, and bounded away from zero in any half-plane $\Re s>-1/2+\epsilon$. By Perron's formula (Theorem 5.2 and Corollary 5.3), we have $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} =\frac{1}{2\pi i}\int_{\sigma_0-iT}^{\sigma_0+iT}F(s)\frac{z^s}{s}dz+R,$$ where $T>0$ is arbitrary, $\sigma_0:=\frac{1}{\log z}$, and $$R \ll \sum_{ z/2 < d < 2z } \frac{\tau(d)}{d} \min\left(1,\frac{z}{T|z-d|}\right) +\frac{1}{T}\sum_d\frac{\tau(d)}{d^{1+\sigma_0}}.$$ It is straightforward to estimate the right hand side to yield $$R \ll z^{-\frac{1}{2}+\epsilon}+\frac{\log^2 z}{T}.$$ This relies on To see this, we estimate the observation first term as $$\sum_{ z/2 < d < 2z } \tau(d) frac{\tau(d)}{d} \min\left(1,|z-d|^{-1}\right) min\left(1,\frac{z}{T|z-d|}\right) \ll z^{-\frac{1}{2}+\epsilon}+\frac{1}{T}\sum_{{z/2 < d < 2z}\atop{|z-d|>\sqrt{2z}}}\frac{\tau(d)}{|z-d|}$$ $$\sum_{ ll z^{-\frac{1}{2}+\epsilon}+\frac{1}{T}\sum_{ \ell < \sqrt{2z}}\frac{1}{\ell}\sum_{z/(2\ell) < m < 2z/\ell} \min\left(1,\left|\frac{z}{\ell}-m\right|^{-1}\right),$$ and applying for the inner sum we apply the argument on page 180 in the book. We shall use $T:=\exp(\sqrt{c\log z})$ for some small absolute constant $c>0$, so that $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} =\frac{1}{2\pi i}\int_{\sigma_0-iT}^{\sigma_0+iT}F(s)\frac{z^s}{s}dz +O\left(\frac{\log^2 z}{T}\right).$$ Using Theorem 6.7 we can see that the integrand is holomorphic in the rectangle with vertices $\sigma_0\pm iT$ and $\sigma_1\pm iT$, for $\sigma_1:=-\frac{c}{\log T}$ and $c>0$ a small absolute constant. Moreover, we can estimate the integrand on the sides of the rectangle. Hence applying Cauchy's theorem and estimates as on page 181 of the book, we obtain $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} \ll z^{-\sigma_1}(\log z)^3+\frac{\log^2 z}{T}.$$ The right hand side is $\ll\exp(-(c/2)\sqrt{\log z})$, proving the original claim.
Let me elaborate on Matt Young's comment and show, for a sufficiently small absolute constant $c>0$, $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} \ll \exp(-c\sqrt{\log z}).$$ All references will be from Montgomery-Vaughan: Multiplicative number theory I.
We can assume that $z>2$ is not an integer. The associated Dirichlet series $$F(s):=\sum_{d=1}^\infty\mu(d)\frac{\tau(d)}{d^{s+1}} =\prod_p\left(1-\frac{2}{p^{s+1}}\right)=\frac{G(s)}{\zeta(s+1)^2}$$ is absolutely convergent in $\Re s>0$, and $G(s)$ is given by an absolutely convergent Euler product in $\Re s>-1/2$. In particular, $G(s)$ is holomorphic, bounded, and bounded away from zero in any half-plane $\Re s>-1/2+\epsilon$. By Perron's formula (Theorem 5.2 and Corollary 5.3), we have $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} =\frac{1}{2\pi i}\int_{\sigma_0-iT}^{\sigma_0+iT}F(s)\frac{z^s}{s}dz+R,$$ where $T>0$ is arbitrary, $\sigma_0:=\frac{1}{\log z}$, and $$R \ll \sum_{ z/2 < d < 2z } \frac{\tau(d)}{d} \min\left(1,\frac{z}{T|z-d|}\right) +\frac{1}{T}\sum_d\frac{\tau(d)}{d^{1+\sigma_0}}.$$ It is straightforward to estimate the right hand side to yield $$R \ll z^{-\frac{1}{2}+\epsilon}+\frac{\log^2 z}{T}.$$ This relies on the observation $$\sum_{ z/2 < d < 2z } \tau(d) \min\left(1,|z-d|^{-1}\right) \ll \sum_{ \ell < \sqrt{2z}}\frac{1}{\ell}\sum_{z/(2\ell) < m < 2z/\ell} \min\left(1,\left|\frac{z}{\ell}-m\right|^{-1}\right),$$ and applying for the inner sum the argument on page 180 in the book. We shall use $T:=\exp(\sqrt{c\log z})$ for some small absolute constant $c>0$, so that $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} =\frac{1}{2\pi i}\int_{\sigma_0-iT}^{\sigma_0+iT}F(s)\frac{z^s}{s}dz +O\left(\frac{\log^2 z}{T}\right).$$ Using Theorem 6.7 we can see that the integrand is holomorphic in the rectangle with vertices $\sigma_0\pm iT$ and $\sigma_1\pm iT$, for $\sigma_1:=-\frac{c}{\log T}$ and $c>0$ a small absolute constant. Moreover, we can estimate the integrand on the sides of the rectangle. Hence applying Cauchy's theorem and estimates as on page 181 of the book, we obtain $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} \ll z^{-\sigma_1}(\log z)^3+\frac{\log^2 z}{T}.$$ The right hand side is $\ll\exp(-(c/2)\sqrt{\log z})$, proving the original claim.