Let $X \sim f_X(x)$ and $Y = g(X)$. If $g(X)$ is a differentiable, monotonic function with inverse such that $X = g^{-1}(Y)$ then the PDF of $Y$ can be described:
$$ f_Y(y) = f_X(g^{-1}(y)) \bigg| \dfrac{d}{dy}g^{-1}(y) \bigg| $$
In the multidimensional case, I have seen a very similar expression where the absolute value of the derivative is replaced by the absolute value of the determinant of the Jacobian:
$f_Y(\pmb{y}) = f_X(g^{-1}(\pmb{y})) |\bigg| \dfrac{\partial \pmb{x}}{\partial \pmb{y}} \bigg| |$
However, I have not seen a qualifier that in the multivariate case the function $\pmb{y} =g(\pmb{x})$ must be monotonic in all variables. Is this assumed or does this requirement not apply in this case? If not why?
The reason is that it is assumed that the system $$\pmb{y} =g(\pmb{x})$$ has a unique solution $x=g^{-1}(\pmb{y})$, i.e., $g$ is invertible when using the formula. In the univariate case, the function $g$ is invertible if and only if $g$ is strictly monotone, which is not the case in the multivariate case ($y_1=x_1, y_2=-x_2$ is a simple example where $g$ is not monotone). Therefore, the condition of strict monotonicity of $g$ used in univariate case is replaced with other equivalent conditions holding in the multivariate case (i.e., the system has a unique solution or $g$ is invertible).
Suppose that the system has multiple solutions:
$$\pmb{x}=h^j(\pmb{y}), j=1,\dots, k$$
such that the union of the sets $h^j(S_Y), j=1,\dots,k$ is $S_X$ with null intersections ($S_X$ and $S_Y$ denote the support of $X$ and $Y$), that is, $h^j(S_Y)\cap h^k(S_Y)$ has Lebesgue measure zero for each $j \neq k$, and $\cup_{j=1}^k h^j(S_Y)=S_X$. Then, the following
$$\color{blue}{f_Y(\pmb{y}) = \sum_{j=1}^kf_X(h^j(\pmb{y})) \left | \det \left (\dfrac{\partial h^j(\pmb{y})}{\partial \pmb{y}} \right ) \right |}$$
is a useful formula. For example, for $y=x^2$, $h^1(y)=\sqrt{x}$ and $h^2(y)=-\sqrt{x}$ with $S_X=\mathbb R$ and $S_Y=\mathbb R_{\, \ge0}$. You can see $$h^1(\mathbb R_{\, \ge0})=\mathbb R_{\, \ge0}, h^2(\mathbb R_{\, \ge0})=\mathbb R_{\, \le0},$$
with null intersection $\{ 0\}$ and union $\mathbb R$. Thus, we have for $y\ge0$:
$$f_Y(y)=\frac{1}{\sqrt{y}}f_X(\sqrt{y}) +\frac{1}{\sqrt{y}}f_X(-\sqrt{y}).$$
For more details and examples, you can see pages 6-10 of this note.