Step 1: For $Au_{xx}+2Bu_{xy}+Cu_{yy}+Du_x+Eu_y=0$ with $A=1+x^2$, $B=0$, $C=1+y^2$, $D=x$, $E=y$, the discriminant is \[B^2-4AC=-4(1+x^2)(1+y^2)<0 \text{ always}\] so the equation is elliptic everywhere and statement (I), which claims it is parabolic, is false.
Step 2: Under a coordinate change $\xi=\xi(x,y)$, $\eta=\eta(x,y)$, the new coefficients follow the general rules \[A^*=A\xi_x^2+2B\xi_x\xi_y+C\xi_y^2, \quad C^*=A\eta_x^2+2B\eta_x\eta_y+C\eta_y^2\] \[B^*=A\xi_x\eta_x+B(\xi_x\eta_y+\xi_y\eta_x)+C\xi_y\eta_y\] \[D^*=A\xi_{xx}+2B\xi_{xy}+C\xi_{yy}+D\xi_x+E\xi_y, \quad E^*=A\eta_{xx}+2B\eta_{xy}+C\eta_{yy}+D\eta_x+E\eta_y\]
Step 3: Choose $\xi=\sinh^{-1}x$ (function of $x$ only) and $\eta=\sinh^{-1}y$ (function of $y$ only). Then \[\xi_x=\frac{1}{\sqrt{1+x^2}},\ \xi_y=0,\ \xi_{xx}=\frac{-x}{(1+x^2)^{3/2}},\ \xi_{xy}=0\] \[\eta_y=\frac{1}{\sqrt{1+y^2}},\ \eta_x=0,\ \eta_{yy}=\frac{-y}{(1+y^2)^{3/2}},\ \eta_{xy}=0\]
Step 4: Substitute into the formulas: \[A^*=(1+x^2)\cdot\frac{1}{1+x^2}=1, \qquad C^*=(1+y^2)\cdot\frac{1}{1+y^2}=1, \qquad B^*=0\] \[D^*=(1+x^2)\left(\frac{-x}{(1+x^2)^{3/2}}\right)+x\cdot\frac{1}{\sqrt{1+x^2}}=\frac{-x}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}=0\] \[E^*=(1+y^2)\left(\frac{-y}{(1+y^2)^{3/2}}\right)+y\cdot\frac{1}{\sqrt{1+y^2}}=0\]
Step 5: All lower-order coefficients vanish, so the canonical form is simply \[A^*u_{\xi\xi}+2B^*u_{\xi\eta}+C^*u_{\eta\eta}+D^*u_\xi+E^*u_\eta=u_{\xi\xi}+u_{\eta\eta}=0\] This confirms statement (II) is true, while (I) is false. \[\boxed{\text{Only (II)}}\]