Let $S$ be the portion of the cylinder $y=\ln(x)$ (what, this is a cylinder?) in the first octant such that the projector parallel to $y$ over the plane $xz$ is the rectangle $1\le x\le e$, $0\le z\le 1$. Let $\vec{n}$ be the unit normal vector to $S$ which points 'outwards' the $xz$ plane. Find the flux of $F=2y\vec{j}+z\vec{k}$ over $S$ in the direction of $\vec{n}$
So, the flux can be computed with:
$$\iint_K \vec{F}(\sigma(x,y))\cdot\vec{n}(\sigma(x,y)||\frac{\partial z}{\partial x}\times \frac{\partial z}{\partial y}|| \ dxdy=$$ $$\iint_K \vec{F}(\sigma(x,y))\cdot\frac{\partial z}{\partial x}\times \frac{\partial z}{\partial y}dxdy$$
But since here I have a function $y$ with respect to $x$ I'm gonna see $y$ as the dependent variable, so our integral uses $y$ in function of $x$ and $z$: $$\iint_K \vec{F}(\sigma(x,z))\cdot\frac{\partial y}{\partial x}\times \frac{\partial y}{\partial z}dxdz$$
I'm seeing $y=\ln(x)$ over the plane $xy$, so its projection is the rectangle $1\le x\le e$, $0\le z\le 1$. The vectors $y$ is:
$$(x,\ln(x),z)$$
The partial derivatives are:
$$\frac{\partial y}{\partial x} = (1,\frac{1}{x},0)$$ $$\frac{\partial y}{\partial z} = (0,0,1)$$
So its cross product is $(\frac{1}{x},-1,0)$ or $(\frac{1}{x},1,0)$ (which is the one we want because it points outwards).
Also, $\vec{F}$ when viewed in terms of $x$ and $z$ becomes:
$$F=2y\vec{j}+z\vec{k} = 2\ln(x)\vec{j}+z\vec{k}$$
So our integral becomes:
$$\iint_K (0,2\ln(x),z)\cdot (\frac{1}{x},1,0) dA = 2\int_0^1\int_0^e \ln(x) \ dxdz = 2\int_0^1 [x(\ln(x)-1)]_0^e \ dz = 0$$
So, the flux is $0$? I can't plot the vector field $\vec{F}$ in Wolfram Alpha to see if it makes sense, but I don't know if I'm right. Could somebody help me?
Using your notation, $$\frac{\partial y}{\partial x}\times \frac{\partial y}{\partial z} = \left(\frac 1 x, -1, 0\right)$$ and $$\frac{\partial y}{\partial z}\times \frac{\partial y}{\partial x} = \left(\color{red}-\frac 1 x, 1, 0\right)$$ may also be used instead (though I am unsure as to how you are defining the "inside" of an open cylinder), but $(1/x, 1, 0)$ is neither one, and is not an appropriate choice. However, since $\vec F$ has no component in the $x$ direction, this error makes no impact.
But, the limits of integration for $x$ are $1$ to $e$, not $0$ to $e$. Also, $$\int \ln x\,dx = x\ln x - x + C,$$ not $x\ln x - 1 + C$. Correct these, and you will see that the flux is not $0$ after all.