<s>[INST] Calculate the minimum value of $T = 3x + 2y$ with positives $x$ and $y$ satisfying $\log_3\frac{xy + 2}{x^2 + 1} = \frac{3^x + 3x^2 + 1}{3^x + xy}$. 
Consider two positives $x$ and $y$ such that $\log_3\dfrac{xy + 2}{x^2 + 1} = \dfrac{3^x + 3x^2 + 1}{3^x + xy}$. Calculate the minimum value of $T = 3x + 2y$.

We have that $$\begin{aligned} \log_3\dfrac{xy + 2}{x^2 + 1} = \dfrac{3^x + 3x^2 + 1}{3^x + xy} \iff \log_3(xy + 2) - \log_3(3x^2 + 3) = \dfrac{(3x^2 + 3) - (xy + 2)}{3^x + xy} \end{aligned}$$
I don't know how to notate this, but let $z$ be a variable and $f(x), g(x)$ be functions such that $g(f(x), z) = f(x) + (3^z + yz)\log_3f(x)$, which means that $g(xy + 2, x) = g(3x^2 + 3, x)$ and $$\dfrac{\mathrm dg(f(x), x)}{\mathrm dx} = f'(x) + \dfrac{(3^x\ln 3 + y)\ln f(x)}{\ln 3} + \dfrac{(3^x + xy)f'(x)}{\ln 3f(x)}$$
Entering the equation $g'(x) = 0$ into WolframAlpha, function $f(x)$ can be written in the form of $$f(x) = \dfrac{3^x + xy}{\ln 3}W\left(\dfrac{\ln 3}{(3^x + xy)e^{(3^x + xy)/c}}\right)$$ with $c$ being a constant and $W(z)$ being the product log function/Lambert $W$ function. Now, I don't know about this function, so I'd consider this a dead end.
I do have a hunch that $xy + 2 = 3x^2 + 3$, I just don't know how to prove it.
How about continuing from this point on? It can be implied that $y = \dfrac{3x^2 + 1}{x}$, which makes $T$ equal to $3x + \dfrac{2(3x^2 + 1)}{x}$. From there, $T' = 0 \iff 9 - \dfrac{2}{x^2} = 0 \iff x = \dfrac{\sqrt 2}{3}$
Drawing the table of variations for $T$, it can be seen that $\min T = 9 \cdot \dfrac{\sqrt 2}{3} + \dfrac{2}{\dfrac{\sqrt 2}{3}} = 6\sqrt{2}$.
Now, if I could just prove that $xy + 2 = 3x^2 + 3$, it would be fantastic. So I turn to your support.
As always, thanks for reading, (and even more if you could help~)
 [/INST] Build the auxiliary Lagrangian function
$$\mathscr L=3x+2y-\lambda\left(\log_3\frac{xy+2}{x^2+1}-\frac{3^x+3x^2+1}{3^x+xy}\right)+\mu_1x+\mu_2y.$$
The first order conditions for a minimum (see KKT multipliers) are
\begin{align}
 \frac{\partial\mathscr L}{\partial x}&=0\\
 \frac{\partial\mathscr L}{\partial y}&=0\\
 \frac{\partial\mathscr L}{\partial \lambda}&=0\\
 \mu_1x&=0\\
 \mu_2y&=0
\end{align}
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