Step 1: Read the two equations.
The first equation is $x^2 - 8xy + 16y^2 = 0$. The second one mixes logarithms of $x$ and $y$. Our plan is to first squeeze a simple link between $x$ and $y$ from the first equation, then plug that into the second.
Step 2: Factor the first equation neatly.
Notice $x^2 - 8xy + 16y^2$ is a perfect square. It equals $(x - 4y)^2$. So $(x - 4y)^2 = 0$, which forces $x = 4y$.
Step 3: Recognise the second equation is also a perfect square.
The expression $(\log_{10} x)^2 + 2(\log_{10} x)(\log_{10} y) + (\log_{10} y)^2$ is just $(\log_{10} x + \log_{10} y)^2$. So $(\log_{10} x + \log_{10} y)^2 = 4$.
Step 4: Turn the log sum into a product.
Since $\log_{10} x + \log_{10} y = \log_{10}(xy)$, we get $(\log_{10}(xy))^2 = 4$, so $\log_{10}(xy) = \pm 2$, meaning $xy = 100$ or $xy = \frac{1}{100}$.
Step 5: Combine $x = 4y$ with these products.
Put $x = 4y$ into $xy = 100$: $4y^2 = 100$, so $y^2 = 25$, giving $y = 5$ and $x = 20$. Put $x = 4y$ into $xy = \frac{1}{100}$: $4y^2 = \frac{1}{100}$, so $y^2 = \frac{1}{400}$, giving $y = \frac{1}{20}$ and $x = \frac{1}{5}$.
Step 6: Match with the options.
The pair $x = \frac{1}{5}, y = \frac{1}{20}$ appears directly in the choices and fits both equations.
\[ \boxed{x = \tfrac{1}{5},\ y = \tfrac{1}{20}} \]