Let $\left(X,Y\right)$ be a two dimensional normal random variable, where $X\overset{d}{=}N\left(0,1\right)$ and $Y\overset{d}{=}N\left(0,1\right)$ with $\operatorname{cov}\left(X,Y\right)=\rho$. How can we prove $$\operatorname{cov}\left(f\left(X\right),g\left(Y\right)\right)\leq\left|\rho\right|\mathbb{D}\left(f\left(X\right)\right)\mathbb{D}\left(g\left(Y\right)\right),$$ where $f\left(X\right),g\left(Y\right)\in L^{2}\left(\Omega\right)$?
This is a homework and we got the following hint: Let $\left(X_{t},Y_{t}\right)_{t}$ a two dimensional Wiener-process, with constant $\rho$ correlation. We should write somehow $f\left(X_{1}\right)-\mathbb{E}\left(f\left(X_{1}\right)\right)$ and $g\left(Y_{1}\right)-\mathbb{E}\left(g\left(Y_{1}\right)\right)$ into integral form, and than we should use the Itô isometry.
Here is what I did so far:
I used Itô's formula for $f\left(X_{t}\right)-\mathbb{E}\left(X_{t}\right)$, and I got $$f\left(X_{1}\right)-\mathbb{E}\left(f\left(X_{1}\right)\right)=\underbrace{f\left(X_{0}\right)-\mathbb{E}\left(f\left(X_{0}\right)\right)}_{=0}+\int_{0}^{1}f'\left(X_{s}\right)dX_{s}+\frac{1}{2}\int_{0}^{1}f''\left(X_{s}\right)ds.$$ I also did it for the other expression, and I got: $$g\left(Y_{1}\right)-\mathbb{E}\left(g\left(Y_{1}\right)\right)=\underbrace{g\left(Y_{0}\right)-\mathbb{E}\left(g\left(Y_{0}\right)\right)}_{=0}+\int_{0}^{1}g'\left(Y_{s}\right)\rho dX_{s}+\int_{0}^{1}g'\left(Y_{s}\right)\sqrt{1-\rho^{2}}dX_{s}^{*}+\frac{1}{2}\int_{0}^{1}g''\left(Y_{s}\right)ds,$$ where $\left(X_{s}\right)_{s}$ and $\left(X_{s}^{*}\right)_{s}$ are independent Wiener processes. By definition, I got the following for covariance: $$\operatorname{cov}\left(f\left(X_{1}\right),g\left(Y_{1}\right)\right) =\mathbb{E}\left(\left[f\left(X_{1}\right)-\mathbb{E}\left(f\left(X_{1}\right)\right)\right]\cdot\left[g\left(Y_{1}\right)-\mathbb{E}\left(g\left(Y_{1}\right)\right)\right]\right)= \mathbb{E}\left(\left[\int_{0}^{1}f'\left(X_{s}\right)dX_{s}+\frac{1}{2}\int_{0}^{1}f''\left(X_{s}\right)ds\right]\cdot\left[\int_{0}^{1}g'\left(Y_{s}\right)\rho dX_{s}+\int_{0}^{1}g'\left(Y_{s}\right)\sqrt{1-\rho^{2}}dX_{s}^{*}+\frac{1}{2}\int_{0}^{1}g''\left(Y_{s}\right)ds\right]\right).$$
Here I want to use somehow the Cauchy–Schwarz inequality: $$\mathbb{E}\left(\left[f\left(X_{1}\right)-\mathbb{E}\left(f\left(X_{1}\right)\right)\right]\cdot\left[g\left(Y_{1}\right)-\mathbb{E}\left(g\left(Y_{1}\right)\right)\right]\right)\leq\underbrace{\sqrt{\mathbb{E}\left(f\left(X_{1}\right)-\mathbb{E}\left(f\left(X_{1}\right)\right)^{2}\right)}}_{\overset{\circ}{=}\mathbb{D}\left(f\left(X_{1}\right)\right)}\cdot\sqrt{\mathbb{E}\left(\left(g\left(Y_{1}\right)-\mathbb{E}\left(g\left(Y_{1}\right)\right)\right)^{2}\right)},$$ and for $\mathbb{E}\left(\left(g\left(Y_{1}\right)-\mathbb{E}\left(g\left(Y_{1}\right)\right)\right)^{2}\right)$ using the Itô isometry:
$$\mathbb{E}\left(\left(g\left(Y_{1}\right)-\mathbb{E}\left(g\left(Y_{1}\right)\right)\right)^{2}\right) =\mathbb{E}\left(\int_{0}^{1}\left(g'\left(Y_{s}\right)\right)^{2}\rho^{2}ds\right)+\mathbb{E}\left(\int_{0}^{1}\left(g'\left(Y_{s}\right)\right)^{2}\left(1-\rho^{2}\right)ds\right)+0 +2\cdot0+2\mathbb{E}\left(\int_{0}^{1}g'\left(Y_{s}\right)\rho dX_{s}\cdot\frac{1}{2}\int_{0}^{1}g''\left(Y_{s}\right)ds\right)+2\mathbb{E}\left(g'\left(Y_{s}\right)\sqrt{1-\rho^{2}}dX_{s}^{*}\cdot\frac{1}{2}\int_{0}^{1}g''\left(Y_{s}\right)ds\right).$$ From this point I don't know how to continue...
Anyway, is it a right direction to prove this statement with the hints?