Let $x$ and $y$ be positive numbers which satisfy $$\left(\log_2 x\right)^2 + \left(\log_2 y\right)^2 = \log_2 \frac{8x^2}{y^2}. \tag{1}$$ We are to find the maximum value of $xy^2$ and the values of $x$ and $y$ at that point. (1) The right side of (1) can be transformed into $$\log_2 \frac{8x^2}{y^2} = \mathbf{A}\log_2 x - \mathbf{B}\log_2 y + \mathbf{C}.$$ So, setting $X = \log_2 x$ and $Y = \log_2 y$, we can express (1) in terms of $X$ and $Y$ as $$(X - \mathbf{D})^2 + (Y + \mathbf{E})^2 = \mathbf{F}.$$ (2) Set $k = \log_2 xy^2$. Using $X$ and $Y$ above, this equality can be transformed into $$X + \mathbf{GG}\,Y - k = 0.$$ If we graph (2) and (3) on a plane with coordinates $(X, Y)$, the graph of (2) is a circle, and the graph of (3) is a straight line. When $k$ is maximized, the graph of (3) is tangent to the graph of (2). Hence, when $k = \mathbf{H}$, $xy^2$ takes the maximum value IJ and in this case $x = \mathbf{K}$ and $y = \mathbf{L}$.
Let $x$ and $y$ be positive numbers which satisfy
$$\left(\log_2 x\right)^2 + \left(\log_2 y\right)^2 = \log_2 \frac{8x^2}{y^2}. \tag{1}$$
We are to find the maximum value of $xy^2$ and the values of $x$ and $y$ at that point.
(1) The right side of (1) can be transformed into
$$\log_2 \frac{8x^2}{y^2} = \mathbf{A}\log_2 x - \mathbf{B}\log_2 y + \mathbf{C}.$$
So, setting $X = \log_2 x$ and $Y = \log_2 y$, we can express (1) in terms of $X$ and $Y$ as
$$(X - \mathbf{D})^2 + (Y + \mathbf{E})^2 = \mathbf{F}.$$
(2) Set $k = \log_2 xy^2$. Using $X$ and $Y$ above, this equality can be transformed into
$$X + \mathbf{GG}\,Y - k = 0.$$
If we graph (2) and (3) on a plane with coordinates $(X, Y)$, the graph of (2) is a circle, and the graph of (3) is a straight line. When $k$ is maximized, the graph of (3) is tangent to the graph of (2). Hence, when $k = \mathbf{H}$, $xy^2$ takes the maximum value IJ and in this case $x = \mathbf{K}$ and $y = \mathbf{L}$.