1. **State the problem:** Evaluate the definite integral $$\int_{e^{2}}^{e^{4}} \frac{1}{\sqrt{x}(1-\sqrt{x})} \, dx.$$\n\n2. **Substitution:** Let $$u = \sqrt{x} = x^{\frac{1}{2}}.$$ Then, $$x = u^{2}$$ and $$dx = 2u \, du.$$\n\n3. **Change the limits:** When $$x = e^{2},$$ $$u = \sqrt{e^{2}} = e.$$ When $$x = e^{4},$$ $$u = \sqrt{e^{4}} = e^{2}.$$\n\n4. **Rewrite the integral:** Substitute into the integral:\n$$\int_{e}^{e^{2}} \frac{1}{u(1-u)} \cdot 2u \, du = \int_{e}^{e^{2}} \frac{2u}{u(1-u)} \, du = \int_{e}^{e^{2}} \frac{2}{1-u} \, du.$$\n\n5. **Simplify the integrand:** The $$u$$ cancels out:\n$$\int_{e}^{e^{2}} \frac{2}{1-u} \, du.$$\n\n6. **Integrate:** Use the formula $$\int \frac{1}{a - u} du = -\ln|a - u| + C.$$ Here, $$a = 1$$, so\n$$\int \frac{2}{1-u} du = 2 \int \frac{1}{1-u} du = -2 \ln|1-u| + C.$$\n\n7. **Evaluate definite integral:**\n$$-2 \ln|1-u| \Big|_{e}^{e^{2}} = -2 \left( \ln|1 - e^{2}| - \ln|1 - e| \right) = -2 \ln \left( \frac{|1 - e^{2}|}{|1 - e|} \right).$$\n\n8. **Simplify the expression:** Since $$e > 1,$$ both $$1 - e$$ and $$1 - e^{2}$$ are negative, so their absolute values are $$e - 1$$ and $$e^{2} - 1$$ respectively. Thus,\n$$-2 \ln \left( \frac{e^{2} - 1}{e - 1} \right).$$\n\n9. **Factor numerator:** Note that $$e^{2} - 1 = (e - 1)(e + 1).$$ So,\n$$-2 \ln \left( \frac{(e - 1)(e + 1)}{e - 1} \right) = -2 \ln (e + 1).$$\n\n**Final answer:** $$\boxed{-2 \ln (e + 1)}.$$