Let $$ p(x, y) \equiv \frac{1}{\sqrt{2\pi}}exp\left(-\frac{x^2}{2}\right)\times\frac{1}{\sqrt{2\pi}}exp\left(-\frac{y^2}{2}\right)$$be the joint standard normal distribution. Let $q(x, y)$ be the density function under Q. Let $L$ be the Lebesgue measure.
Then by definition
$$\frac{dP}{dL}=p(x, y)\quad\text{and}\quad \frac{dQ}{dL}=q(x, y).$$
Hence $$ q(x, y) = \frac{dQ}{dL} = \frac{dQ}{dP}\frac{dP}{dL} = \frac{e^{xy\theta}}{Ee^{XY\theta}} p(x, y).$$
One can verify $e^{xy\theta}$ is not product-term separable function for $\theta \in (0, 1)$. Further note $p(x, y)$ is product-term separable and $Ee^{XY\theta}$ is a function of $\theta$ (considered a constant). Hence $q(x, y)$ is not product-term separable. Hence $X, Y$ are not independent under probability measure $Q$.
PS: Noting the moment generation function for a standard normal distribution is $M(t)=e^{\frac{1}{2}t^2}$,
$$Ee^{XY\theta}=E_XE_Ye^{(X\theta)Y}=E_Xe^{\frac{1}{2}\theta^2X^2} = \int \frac{1}{\sqrt{2\pi}}exp\left(-\frac{(1-\theta^2)x^2}{2}\right)=\frac{1}{\sqrt{1-\theta^2}}.$$
For separability, one can refer to:theorem 3. If you want to avoid separability, you can calculate the marginal densities and show the joint density is not a product of marginals.
