I see that nobody has replied yet, so I'll offer this answer, but I know that there are more qualified experts around than me. So possibly a better answer will appear in due course.

There's a bit more detail in D Bressanini and PJ Reynolds Adv Chem Phys, 105, 37 (1999) eqns (4.10) and (4.11), and in Bryan Clark's Boulder Summer School notes, p5. I don't think that it is possible to derive this formula without assuming that the trial wavefunction is real, making the local energy $E_L(X)\equiv \Psi(X)^{-1} H\Psi(X)$ also real. But maybe I missed something.

Let's follow this through, to highlight the key step. I'll use $\Psi_\alpha$ to denote $\partial\Psi/\partial\alpha$. Differentiating, using the chain rule, $$ \frac{\partial}{\partial\alpha} \langle H\rangle = \frac{\int \text{d}X\, \Psi_\alpha^* H \Psi + \Psi^* H\Psi_\alpha}{\int \text{d}X\, \Psi^* \Psi } - \frac{\left(\int \text{d}X\, \Psi^* H \Psi\right)\left(\int \text{d}X\, \Psi_\alpha^* \Psi + \Psi^*\Psi_\alpha\right)}{\left(\int \text{d}X\, \Psi^* \Psi \right)^2} . $$ The second term is not a problem. We can recognize $\frac{\partial}{\partial\alpha}|\Psi|^2$ inside the last integral of the numerator; multiplying and dividing by $\Psi^* \Psi$ allows us to write it as $$ \frac{\partial}{\partial\alpha} \langle H\rangle = \frac{\int \text{d}X\, \Psi_\alpha^* H \Psi + \Psi^* H\Psi_\alpha}{\int \text{d}X\, \Psi^* \Psi } - \left\langle E_L\right\rangle \left\langle \frac{\partial}{\partial\alpha} \ln |\Psi|^2\right\rangle . $$ For the first term we use the Hermitian property of $H$ to rewrite $\int \text{d}X\, \Psi^* H\Psi_\alpha=\int \text{d}X\, \Psi_\alpha (H\Psi)^*$, so $$ \frac{\partial}{\partial\alpha} \langle H\rangle = \frac{\int \text{d}X\, \Psi_\alpha^* H \Psi + \Psi_\alpha (H\Psi)^*}{\int \text{d}X\, \Psi^* \Psi } - \left\langle E_L\right\rangle \left\langle \frac{\partial}{\partial\alpha} \ln |\Psi|^2\right\rangle . $$ Again, multiply and divide both terms in the numerator of the first integral by $\Psi^* \Psi$ to give \begin{align*} \frac{\partial}{\partial\alpha} \langle H\rangle &= \left\langle \frac{\Psi_\alpha^*}{\Psi^*} \frac{H \Psi}{\Psi}+ \frac{\Psi_\alpha}{\Psi} \left(\frac{H\Psi}{\Psi}\right)^*\right\rangle - \left\langle E_L\right\rangle \left\langle \frac{\partial}{\partial\alpha} \ln |\Psi|^2\right\rangle \\ &= \left\langle \frac{\Psi_\alpha^*}{\Psi^*} E_L+ \frac{\Psi_\alpha}{\Psi} E_L^*\right\rangle - \left\langle E_L\right\rangle \left\langle \frac{\partial}{\partial\alpha} \ln |\Psi|^2\right\rangle . \end{align*} At this point, I think we have to take $\Psi$ to be real, so both the terms in the first bracket are the same, and we can replace $|\Psi|^2$ by $\Psi^2$ in the last term, giving the result you wanted to prove.

I see that nobody has replied yet, so I'll offer this answer, but I know that there are more qualified experts around than me. So possibly a better answer will appear in due course.

There's a bit more detail in D Bressanini and PJ Reynolds Adv Chem Phys, 105, 37 (1999) eqns (4.10) and (4.11), and in Bryan Clark's Boulder Summer School notes, p5. I don't think that it is possible to derive this formula without assuming that the trial wavefunction is real, making the local energy $E_L(X)\equiv \Psi(X)^{-1} H\Psi(X)$ also real. But maybe I missed something.

Let's follow this through, to highlight the key step. I'll use $\Psi_\alpha$ to denote $\partial\Psi/\partial\alpha$. Differentiating, using the chain rule, $$ \frac{\partial}{\partial\alpha} \langle H\rangle = \frac{\int \text{d}X\, \Psi_\alpha^* H \Psi + \Psi^* H\Psi_\alpha}{\int \text{d}X\, \Psi^* \Psi } - \frac{\left(\int \text{d}X\, \Psi^* H \Psi\right)\left(\int \text{d}X\, \Psi_\alpha^* \Psi + \Psi^*\Psi_\alpha\right)}{\left(\int \text{d}X\, \Psi^* \Psi \right)^2} . $$ The second term can be tidied up. We can recognize $\frac{\partial}{\partial\alpha}|\Psi|^2$ inside the last integral of the numerator; multiplying and dividing inside both the integrals of the numerator of the second term by $\Psi^* \Psi$ gives us $$ \frac{\partial}{\partial\alpha} \langle H\rangle = \frac{\int \text{d}X\, \Psi_\alpha^* H \Psi + \Psi^* H\Psi_\alpha}{\int \text{d}X\, \Psi^* \Psi } - \left\langle E_L\right\rangle \left\langle \frac{\partial}{\partial\alpha} \ln |\Psi|^2\right\rangle . $$ For the first term we use the Hermitian property of $H$ to rewrite $\int \text{d}X\, \Psi^* H\Psi_\alpha=\int \text{d}X\, \Psi_\alpha (H\Psi)^*$, so $$ \frac{\partial}{\partial\alpha} \langle H\rangle = \frac{\int \text{d}X\, \Psi_\alpha^* H \Psi + \Psi_\alpha (H\Psi)^*}{\int \text{d}X\, \Psi^* \Psi } - \left\langle E_L\right\rangle \left\langle \frac{\partial}{\partial\alpha} \ln |\Psi|^2\right\rangle . $$ Again, multiply and divide both terms within the integral in the numerator of the first term by $\Psi^* \Psi$ to give \begin{align*} \frac{\partial}{\partial\alpha} \langle H\rangle &= \left\langle \frac{\Psi_\alpha^*}{\Psi^*} \frac{H \Psi}{\Psi}+ \frac{\Psi_\alpha}{\Psi} \left(\frac{H\Psi}{\Psi}\right)^*\right\rangle - \left\langle E_L\right\rangle \left\langle \frac{\partial}{\partial\alpha} \ln |\Psi|^2\right\rangle \\ &= \left\langle \frac{\Psi_\alpha^*}{\Psi^*} E_L+ \frac{\Psi_\alpha}{\Psi} E_L^*\right\rangle - \left\langle E_L\right\rangle \left\langle \frac{\partial}{\partial\alpha} \ln |\Psi|^2\right\rangle . \end{align*} At this point, I think we have to take $\Psi$ to be real, so both the terms in the first bracket are the same, and we can replace $|\Psi|^2$ by $\Psi^2$ in the last term, giving the result you wanted to prove. Obviously, we expect all the final expectation values here to be real, and they manifestly are, so if we don't assume $\Psi$ is real, I think this last equation is the result.