If you do in fact mean [tex]f(1)=f(6)=0[/tex] (as opposed to one of these being the derivative of [tex]f[/tex] at some point), then integrating twice gives
[tex]f''(x) = -\dfrac1{x^2}[/tex]
[tex]f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1[/tex]
[tex]f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2[/tex]
From the initial conditions, we find
[tex]f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0[/tex]
[tex]f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)[/tex]
Eliminating [tex]C_2[/tex], we get
[tex](C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))[/tex]
[tex]-5C_1 = \ln(6)[/tex]
[tex]C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)[/tex]
Then
[tex]\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}[/tex]
