Inspired by Calvin Lin

Write $g(z) = f(100(1+z)).$ Then $g^{(101)}(z) = 100^{101}f^{(101)}(100(1+z))$ by the Chain Rule, so $g^{(101)}(0) = 100^{101}f^{(101)}(100).$

Now $g(z) = (\ln(100(1+z)))^2 = (\ln(100) + \ln(1+z))^2,$ and we can expand the inside as a Taylor series around $z=0$ : $g(z) = \left(\ln(100) + z - \frac{z^2}2 + \frac{z^3}3 - \cdots + \frac{z^{101}}{101} + \cdots\right)^2,$ so multiplying out, we get that the coefficient of $z^{101}$ is $\frac{2\ln(100)}{101} - 2\left( \frac1{1\cdot 100} + \frac1{2\cdot 99} + \cdots + \frac1{50\cdot 51} \right).$ Of course, $g^{(101)}(0)$ is the coefficient of $z^{101}$ times $101!.$

Let's put that all together: $\begin{aligned} f^{(101)}(100) &= \frac{g^{(101)}(0)}{100^{101}} \\ &= \frac{101! \left( \text{coefficient of } z^{101} \right)}{100^{101}} \\ &= \frac{101! \left( \frac{2\ln(100)}{101} - 2\left( \frac1{1\cdot 100} + \frac1{2\cdot 99} + \cdots + \frac1{50\cdot 51} \right)\right)}{100^{101}} \\ &= \frac{100! \left( 2\ln(100) - 2\left( \frac{101}{1\cdot 100} + \frac{101}{2\cdot 99} + \cdots + \frac{101}{50\cdot 51} \right)\right)}{100^{101}} \\ &= \frac{100! \left( 2\ln(100) - 2\left( \frac11 + \frac1{100} + \frac12 + \frac1{99} + \cdots + \frac1{50} + \frac1{51} \right)\right)}{100^{101}} \\ &= -2 \cdot \frac{100!(H_{100} - \ln(100))}{100^{101}}, \end{aligned}$ where $H_n$ is the $n$ th harmonic number . It is well-known that $\lim\limits_{n\to\infty} (H_n - \ln(n)) = \gamma,$ so the answer is $-2 \cdot \frac{100!\gamma}{100^{101}}.$