= Solution
With the printed assumptions alone, the general <quadratic covariation> calculation from the two noise coefficients gives
$$
\begin{aligned}
d[X]&=(X^2+Y^2)\,dt-2XY\,dC,\\
d[Y]&=(X^2+Y^2)\,dt+2XY\,dC,\\
d[X,Y]&=(X^2-Y^2)\,dC.
\end{aligned}
$$
A continuous <finite-variation process> contributes no <quadratic covariation>, so the drift terms in the previous part do not alter these formulas. Taking $\vartheta=B$ gives $d[X,Y]=e^{2B}\cos(2B)dt$, which is nonzero near zero since $B_0=0$. Also $d([X]-[Y])=-4XYdt$ is not identically zero. Thus the requested assertions are false as printed.
Under the corrected assumption $C=0$, $X^2+Y^2=e^{2B}$ yields
$$
\boxed{[X]_t=[Y]_t=A_t:=\int_0^te^{2B_s}\,ds,\qquad [X,Y]_t=0.}
$$
The pair is therefore a pair of <orthogonal continuous local martingales>. This equality also identifies the clock for the intended <time change of a continuous process>.
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