Let $f,g:\mathbb{R} \to \mathbb{R}$ be monotonic and differentiable functions, and suppose that $(X, Y)$ is a random vector such that
$$ (f(X),g(Y))^T \sim N (\mu, \Sigma), $$ where $\mu = (\mu_1, \mu_2)^T$ and $\Sigma$ is a 2x2 covariance matrix. We therefore have that
$$ g(Y)|f(X) \sim N(\mu_2 + \frac{\sigma_{21}}{\sigma_1^2} (f(X) - \mu_1), \sigma^2_2 - \frac{\sigma_{21}}{\sigma_1^2}). $$
I'm interested in the distribution of $Y|X$ though. Is there a closed form expression?
As mentioned and worked out by rumathe in the comments (I had to make a small correction of the last factor), an explicit formula for the probability density of $Y|X$ can be computed using the change of variables formula (for probability densities) under the additional assumption that $f$ and $g$ are diffeomorphisms:
If $f$ is a diffeomorphism, then knowing $X$ and $f(X)$ is equivalent, so $$ g(Y)|X \sim N\bigg(\mu_2 + \frac{\sigma_{21}}{\sigma_1^2} (f(X) - \mu_1), \sigma^2_2 - \frac{\sigma_{21}}{\sigma_1^2}\bigg). $$ To go from the probability density of $g(Y)$ (conditioned on $X$) to the one of $Y$ (conditioned on $X$), one can use the change of variables formula for probability densities applied to $g^{-1}$:
If $Y = h(Z)$ and $h$ is a diffeomorphism, then $$ p_Y(y) = p_Z(h^{-1}(y)) \, \bigg| \frac{\mathrm d h^{-1}}{\mathrm dy} (y)\bigg|. $$ Hence, in your case with $Z = g(Y)$ (conditioned on $X$) and $h = g^{-1}$, $$ p_Y(y|X=x) = \frac{1}{\sqrt{2 \pi (\sigma^2_2 - \frac{\sigma_{21}^2}{\sigma_1^2})}} \, \exp\left(-\frac{(g(y)-\mu_2 - \frac{\sigma_{21}}{\sigma_1^2} (f(x) - \mu_1))^2}{2(\sigma^2_2 - \frac{\sigma_{21}^2}{\sigma_1^2})}\right) \cdot \bigg| \frac{\mathrm d g}{\mathrm dy} (y)\bigg|. $$