Contradiction: finding potential function of $\nabla f$ where $f=e^x y z$ due to $e^0=1$ - violates path independence

I have got the following contradiction:

I take the function $$f(x,y,z)=e^x yz$$ and compute its gradient $$\nabla f=\langle e^xyz,e^x z,e^x y \rangle .$$ Now I want to find the potential (i.e. $f$) from my gradient field again. Actually I want to find $f(x_1,y_1,z_1)$ which should lead to $f(x_1,y_1,z_1)=e^{x_1}y_1z_1$. I choose to integrate first along the $z$-axis, then along the $y$-axis and finally along the $x$-axis. I get:

On the first part $x=0,y=0,dx=0,dy=0.$

$$\int_0^{z_1}e^0 \cdot0 \, dz=0$$

On the second part $x=0, dx=0,z=z_1,dz=0.$

$$\int_0^{y_1}e^0 z_1 \, dy=z_1y_1$$

On the third part $z=z_1,dz=0,y=y_1,dy=0$

$$\int_0^{x_1}e^x y_1z_1 \, dx=e^{x_1}y_1z_1$$

Adding all the results I get $$f(x_1,y_1,z_1)=e^{x_1}y_1z_1 + y_1z_1$$ which is not what I wanted to get. What did go wrong? Since I am integrating over a gradient field I should have path independence. What did I do wrong with $e^0$?

Your final integral should have resulted in $e^{x_1}y_1z_1-y_1z_1$ since $e^0=1$ not $0$, now sum them up and you get the right result.