If we have some closed form multivariable function say 2-in, 1-out, the cross sections of the graph parallel to the $xz$ and $yz$ plane have equations that are inherently closed form.
But is the same true for the diagonal cross sections? Say for example we have the function $f$: $$f(x, y) = e^y\sin(x)$$ If we're given a $\vec{v}$ say, [1, 2] would the cross section of the graph of the plane formed by the $z$-axis and $\vec{v}$ inherently be closed form as $f$ is on the axes?
The answer is yes, and it's easy to see why. Let's say we have some "nice" expression for $f(x,y)$. We want to consider the graph made by slicing the plane by some arbitrary vector, say $\mathbf{v}\in\mathbb{R}^2$. More precisely, what we are doing here is considering a the $z$ coordinate of a path $$\mathbf r(t)=\bigg(x(t)~,~y(t)~,~f\big(x(t),y(t)\big)\bigg)$$ Where $x(t),y(t)$ are just straight lines given by the components of $\mathbf{v}$: $$x(t)=v_xt$$ $$y(t)=v_yt$$
So all we have to do is plug in these expressions to get our graph. In the case of $f(x,y)=\sin(x)e^y$ and $\mathbf{v}=(1,2)$ this is simply $$\sin(t)e^{2t}$$ I suggest you check this graphically to confirm.