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Note that $$ \psi(m,x) =(-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(x+k)^{m+1}}. $$ Therefore $$ \psi(m,1/6) = (-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(k+1/6)^{m+1}} =(-1)^{m+1} m! 6^{m+1} \sum_{n\equiv 1 \mod 6} \frac{1}{n^{m+1}}. $$ Writing the condition $n\equiv 1 \mod 6$ as $n\equiv 1 \mod 3$ but not $4 \mod 6$, the above is \begin{align*} &(-1)^{m+1} m! 6^{m+1} \Big( \sum_{n\equiv 1 \mod 3} \frac{1}{n^{m+1}} - \frac{1}{2^{m+1}} \sum_{n\equiv 2 \mod 3} \frac{1}{n^{m+1}}\Big)\\ &= 2^{m+1} \psi(m,1/3)-\psi(m,2/3). \end{align*} We also have $$ \psi(m,1/3) +\psi(m,2/3) = (-1)^{m+1} m! 3^{m+1} \sum_{n \not\equiv 0\mod 3} \frac{1}{n^{m+1}} = (-1)^{m+1} m! (3^{m+1} -1) \zeta(m+1). $$ From these two relations, clearly we have a linear relation connecting $\psi(m,1/6)$, $\psi(m,1/3)$ and $\zeta(m+1)$.: namely, $$ \psi(m,1/6) = (2^{m+1}+1) \psi(m,1/3)+(-1)^m m! (3^{m+1}-1) \zeta(m+1). $$

Note that $$ \psi(m,x) =(-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(x+k)^{m+1}}. $$ Therefore $$ \psi(m,1/6) = (-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(k+1/6)^{m+1}} =(-1)^{m+1} m! 6^{m+1} \sum_{n\equiv 1 \mod 6} \frac{1}{n^{m+1}}. $$ Writing the condition $n\equiv 1 \mod 6$ as $n\equiv 1 \mod 3$ but not $4 \mod 6$, the above is \begin{align*} &(-1)^{m+1} m! 6^{m+1} \Big( \sum_{n\equiv 1 \mod 3} \frac{1}{n^{m+1}} - \frac{1}{2^{m+1}} \sum_{n\equiv 2 \mod 3} \frac{1}{n^{m+1}}\Big)\\ &= 2^{m+1} \psi(m,1/3)-\psi(m,2/3). \end{align*} We also have $$ \psi(m,1/3) +\psi(m,2/3) = (-1)^{m+1} m! 3^{m+1} \sum_{n \not\equiv 0\mod 3} \frac{1}{n^{m+1}} = (-1)^{m+1} m! (3^{m+1} -1) \zeta(m+1). $$ From these two relations, clearly we have a linear relation connecting $\psi(m,1/6)$, $\psi(m,1/3)$ and $\zeta(m+1)$.

Note that $$ \psi(m,x) =(-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(x+k)^{m+1}}. $$ Therefore $$ \psi(m,1/6) = (-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(k+1/6)^{m+1}} =(-1)^{m+1} m! 6^{m+1} \sum_{n\equiv 1 \mod 6} \frac{1}{n^{m+1}}. $$ Writing the condition $n\equiv 1 \mod 6$ as $n\equiv 1 \mod 3$ but not $4 \mod 6$, the above is \begin{align*} &(-1)^{m+1} m! 6^{m+1} \Big( \sum_{n\equiv 1 \mod 3} \frac{1}{n^{m+1}} - \frac{1}{2^{m+1}} \sum_{n\equiv 2 \mod 3} \frac{1}{n^{m+1}}\Big)\\ &= 2^{m+1} \psi(m,1/3)-\psi(m,2/3). \end{align*} We also have $$ \psi(m,1/3) +\psi(m,2/3) = (-1)^{m+1} m! 3^{m+1} \sum_{n \not\equiv 0\mod 3} \frac{1}{n^{m+1}} = (-1)^{m+1} m! (3^{m+1} -1) \zeta(m+1). $$ From these two relations, clearly we have a linear relation connecting $\psi(m,1/6)$, $\psi(m,1/3)$ and $\zeta(m+1)$: namely, $$ \psi(m,1/6) = (2^{m+1}+1) \psi(m,1/3)+(-1)^m m! (3^{m+1}-1) \zeta(m+1). $$

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Note that $$ \psi(m,x) =(-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(x+k)^{m+1}}. $$ Therefore $$ \psi(m,1/6) = (-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(k+1/6)^{m+1}} =(-1)^{m+1} m! 6^{m+1} \sum_{n\equiv 1 \mod 6} \frac{1}{n^{m+1}}. $$ Writing the condition $n\equiv 1 \mod 6$ as $n\equiv 1 \mod 3$ but not $4 \mod 6$, the above is \begin{align*} &(-1)^{m+1} m! 6^{m+1} \Big( \sum_{n\equiv 1 \mod 3} \frac{1}{n^{m+1}} - \frac{1}{2^{m+1}} \sum_{n\equiv 2 \mod 3} \frac{1}{n^{m+1}}\Big)\\ &= 2^{m+1} \psi(m,1/3)-\psi(m,2/3). \end{align*} We also have $$ \psi(m,1/3) +\psi(m,2/3) = (-1)^{m+1} m! 3^{m+1} \sum_{n \not\equiv 0\mod 3} \frac{1}{n^{m+1}} = (-1)^{m+1} m! (3^{m+1} -1) \zeta(m+1). $$ From these two relations, clearly we have a linear relation connecting $\psi(m,1/6)$, $\psi(m,1/3)$ and $\zeta(m+1)$.