Put $$x = e^{y} \;(\Rightarrow \;y = \ln x,\;dx = e^{y}\,dy)$$.
When $$x$$ moves from $$e^{-3}$$ to $$e^{3}$$, $$y$$ moves from $$-3$$ to $$3$$.
Because $$dx = e^{y}dy$$ and the integrand contains the factor $$1/x$$, this $$e^{y}$$ cancels:
$$I = \int_{-3}^{3}\frac{\bigl[f(e^{|y|}) + f(e^{3-|y|})\bigr]\,
\ln(1+y^{2})}{\ln(1+y^{2})+\ln\!\bigl(1+(3-|y|)^{2}\bigr)}\,dy$$
The integrand is an even function of $$y$$, so
$$I = 2\int_{0}^{3}\frac{\bigl[f(e^{y})+f(e^{3-y})\bigr]\,
\ln(1+y^{2})}{\ln(1+y^{2})+\ln\!\bigl(1+(3-y)^{2}\bigr)}\,dy$$
For brevity, set
$$a(y)=\ln(1+y^{2}),\qquad
b(y)=\ln\!\bigl(1+(3-y)^{2}\bigr)\;.$$
Then
$$I = 2\int_{0}^{3}\Bigl[f(e^{y})+f(e^{3-y})\Bigr]\,
\frac{a(y)}{a(y)+b(y)}\,dy\;.$$
Split the bracket and treat the two pieces separately:
$$I = 2\int_{0}^{3}f(e^{y})\frac{a}{a+b}\,dy
\;+\; 2\int_{0}^{3}f(e^{3-y})\frac{a}{a+b}\,dy\;.\qquad -(1)$$
In the second integral of (1) use the change of variable $$t = 3-y$$
($$dt = -dy$$, limits again $$0 \to 3$$).
Notice that under this change $$a(y)=b(t)$$ and $$b(y)=a(t)$$.
Therefore
$$\int_{0}^{3}f(e^{3-y})\frac{a}{a+b}\,dy
= \int_{0}^{3}f(e^{t})\frac{b}{a+b}\,dt\;.$$
Substituting back into (1) gives
$$\frac{I}{2}= \int_{0}^{3}f(e^{y})\frac{a}{a+b}\,dy
+ \int_{0}^{3}f(e^{y})\frac{b}{a+b}\,dy
= \int_{0}^{3}f(e^{y})\,dy\;,$$
because $$\dfrac{a}{a+b}+\dfrac{b}{a+b}=1$$ for every $$y$$.
Hence
$$I = 2\int_{0}^{3}f(e^{y})\,dy = 2\int_{0}^{3}f\!\left(e^{x}\right)\,dx\;.$$
Therefore the integral equals $$2\int_{0}^{3}f\!\left(e^{x}\right)dx$$, i.e. Option A.
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