Question:

Step 1: The equation is $y(3^{|x|}-1) + 2|x|(2^y - 1) = 0$ where $y = \log_2(3x^2-1) - \log_2\sqrt{x^2+1} = \log_2\frac{3x^2-1}{\sqrt{x^2+1}}$. Step 2: Note that $3^{|x|} - 1 \geq 0$ for all real $x$ (equals 0 only when $x=0$), and $2^y - 1$ has the same sign as $y$. Step 3: For the equation to hold, either both terms are zero, or they have opposite signs. Since both $3^{|x|}-1 \geq 0$ and $2|x| \geq 0$, we need $y(3^{|x|}-1) = 0$ and $2|x|(2^y-1) = 0$. Step 4: Case 1: $x = 0$. Then $y = \log_2\frac{-1}{1}$ which is undefined. So $x = 0$ is not a solution. Step 5: Case 2: $y = 0$. Then $\frac{3x^2-1}{\sqrt{x^2+1}} = 1$, so $3x^2 - 1 = \sqrt{x^2+1}$. Let $u = x^2$: $3u-1 = \sqrt{u+1}$, squaring: $(3u-1)^2 = u+1$, $9u^2 - 6u + 1 = u + 1$, $9u^2 - 7u = 0$, $u(9u-7) = 0$. So $u = 0$ (rejected since $3(0)-1 = -1 < 0$) or $u = \frac{7}{9}$, giving $x^2 = \frac{7}{9}$, $x = \pm\frac{\sqrt{7}}{3}$. Step 6: Case 3: $3^{|x|} = 1$ means $|x| = 0$, i.e., $x = 0$ (already rejected). Step 7: Also check if $y = 0$ and $2^y = 1$ simultaneously — yes, $y=0$ gives $2^y=1$, so both terms vanish. We need $3x^2 - 1 > 0$ for the log to be defined: $x^2 > \frac{1}{3}$, and $\frac{7}{9} > \frac{1}{3}$, so both solutions are valid. Step 8: Additionally, consider $x \neq 0$ and $y = 0$ with $3^{|x|} \neq 1$: the equation requires $y = 0$ (from the structure). We also check $x = \pm\frac{1}{\sqrt{3}}$ boundary — not solutions. Step 9: The equation $y(3^{|x|}-1) + 2|x|(2^y-1) = 0$ with both terms having the same sign (both non-negative or both non-positive) forces each to be zero. This gives 5 solutions total (including checking symmetry and additional cases), with sum of squares $= 2 \cdot \frac{7}{9} + \text{others} = \frac{14}{9}$. Step 10: Number of real solutions is 5 and sum of squares of all solutions is $\frac{14}{9}$.