<s>[INST] Calculating $3^{m-n}=?$ $$9^m + 9^n = 52$$
$$9^m -4 = 2 \cdot 9^n$$
$$3^{m-n}=?$$
Let me show what I've tried

Simpifyling the both equalities. 

$$3^{2^m} + 3^{2n} = 2 \cdot 13 \tag{1}$$
$$3^{2m} -2^2 = 2 \cdot 3^{2n} \tag{2}$$

Diving the second equality by $2$ and we have

$$\frac{3^{2m} -2^2}{2} =3^{2n} \tag{3}$$
Here is where I'm stuck. 
My Kindest Regards!
 [/INST] Putting
$$
x=9^{m}, \quad y=9^{n},
$$
we have
\begin{eqnarray*}
x+y&=&52\\
x-2y&=&4
\end{eqnarray*}
It follows that
$$
9^{m}=x=\frac{3x}{3}=\frac{108}{3}=36, \quad 9^{n}=y=\frac{3y}{3}=\frac{48}{3}=16,
$$
i.e.
$$
3^{m}=\sqrt{9^{m}}=\sqrt{36}=6,\quad 3^{n}=\sqrt{9^{n}}=\sqrt{16}=4.
$$
Hence
$$
3^{m-n}=\frac{3^{m}}{3^{n}}=\frac{6}{4}=\frac{3}{2}.
$$
Remark: The numbers $m$ and $n$ are obviously not integers, in fact 
$$
m=\log_3(6)=\frac{\ln(6)}{\ln(3)}\approx 1.63092975\ldots ,\quad n=\log_3(4)=\frac{\ln(4)}{\ln(3)}\approx 1.261859507\ldots
$$
</s>