1. **State the problem:** Given $x>0$, $y>0$, and the equation $$y^3 \cdot x^{\log_{10}(x)} = 10,$$ find the maximum value of $$x^2 y.$$\n\n2. **Rewrite the constraint:** Note that $\log_{10}(x)$ is the logarithm base 10 of $x$. The constraint is $$y^3 \cdot x^{\log_{10}(x)} = 10.$$\n\n3. **Express $y$ in terms of $x$:** From the constraint, $$y^3 = \frac{10}{x^{\log_{10}(x)}}.$$ Taking cube root, $$y = \left(\frac{10}{x^{\log_{10}(x)}}\right)^{\frac{1}{3}} = 10^{\frac{1}{3}} \cdot x^{-\frac{\log_{10}(x)}{3}}.$$\n\n4. **Objective function:** We want to maximize $$f(x) = x^2 y = x^2 \cdot 10^{\frac{1}{3}} \cdot x^{-\frac{\log_{10}(x)}{3}} = 10^{\frac{1}{3}} \cdot x^{2 - \frac{\log_{10}(x)}{3}}.$$\n\n5. **Simplify the exponent:** Let $$t = \log_{10}(x).$$ Then $$x = 10^t.$$ Substitute into the exponent: $$2 - \frac{t}{3}.$$ So, $$f(t) = 10^{\frac{1}{3}} \cdot (10^t)^{2 - \frac{t}{3}} = 10^{\frac{1}{3}} \cdot 10^{t(2 - \frac{t}{3})} = 10^{\frac{1}{3} + 2t - \frac{t^2}{3}}.$$\n\n6. **Maximize $f(t)$:** Since $10^{\text{something}}$ is increasing, maximize the exponent $$g(t) = \frac{1}{3} + 2t - \frac{t^2}{3}.$$\n\n7. **Find critical points:** Differentiate $$g(t)$$ with respect to $$t$$: $$g'(t) = 2 - \frac{2t}{3}.$$ Set $$g'(t) = 0$$ to find maxima: $$2 - \frac{2t}{3} = 0 \Rightarrow \frac{2t}{3} = 2 \Rightarrow t = 3.$$\n\n8. **Check second derivative:** $$g''(t) = -\frac{2}{3} < 0,$$ so $$t=3$$ is a maximum point.\n\n9. **Calculate maximum value:** Substitute $$t=3$$ back: $$g(3) = \frac{1}{3} + 2(3) - \frac{9}{3} = \frac{1}{3} + 6 - 3 = \frac{1}{3} + 3 = \frac{10}{3}.$$\n\nTherefore, $$f_{max} = 10^{\frac{10}{3}}.$$\n\n10. **Find corresponding $x$ and $y$:** $$x = 10^t = 10^3 = 1000.$$\nFrom step 3, $$y = 10^{\frac{1}{3}} \cdot x^{-\frac{t}{3}} = 10^{\frac{1}{3}} \cdot 10^{-\frac{3}{3}} = 10^{\frac{1}{3} - 1} = 10^{-\frac{2}{3}}.$$\n\n**Final answer:** The maximum value of $$x^2 y$$ is $$\boxed{10^{\frac{10}{3}}}$$ when $$x=1000$$ and $$y=10^{-\frac{2}{3}}.$$