Solve in $\mathbb{R}^2$ the system of equations:
\begin{aligned} 3^x - \frac{1}{y^2} &= 25 \\\\ \log_9(x) - \log_2(y) &= 1 \end{aligned}
We can rewrite the second equation as $\log_3(x) + \log_2\left(\frac{1}{y^2}\right) = 2$ and we can see that $(3, \frac{\sqrt{2}}{2})$ is a solution to the system but I don’t know how to obtain it
By solving your second equation you obtain $$ \frac{1}{y^2}=2^{2-\log_3(x)}=4 \cdot 2^{-\log_3(x)}=4x^{-\frac{1}{\log_2(3)}} $$ So the first equation becames $$ 3^x-4x^{-\frac{1}{\log_2(3)}}=25 $$ Let $$ f(x)=3^x-4x^{-\frac{1}{\log_2(3)}}-25 $$ The derivative is $$ f'(x)=\frac{1}{\log(3)} 3^x+\frac{4}{\log_2(3)} x^{-\frac{1}{\log_2(3)}-1}>0 $$ for each $x>0$ as it is sum of positive quatities. So the function is increasing i.e. it is injective. So the first equation has at most $1$ solution.
As you have checked that $(3, \frac{\sqrt{2}}{2})$ is a solution we must have $x=3$. By plugging the value in the orginal second equation we see $$ \log_2 \frac{1}{y^2}=1 $$ that implies indeed that $y=\pm \frac{\sqrt{2}}{2}$. As the domain of the original system of equation is $x>0, y>0$ we discard the negative solution