How do we plot this implicit function $x+y+z = e^z$?

Question

How do we plot the implicit function $z=f(x,y)$
which is defined by the equation $x+y+z = e^z$?

I kind of know WolframAlpha and SymPy but I am open to other suggestions too.

I am curious to see how this function looks like.
Because for all four 2nd order derivatives I am getting
the same expression $$\frac{-e^z}{(e^z-1)^3}$$
That fact seems quite curious to me.

And for the two 1st order derivatives I got this expression

$$\frac{1}{e^z-1}$$

But this expression is not defined when $z=0$.

I found that $z$ is zero exactly when $x+y=1$.

So I wonder what happens with $z$ when we are on the line $x+y=1$.
Seems $z$ is zero there but OK ... do we really not have $dz/dx$ and $dz/dy$ when we are on that line?!

All in all this $z$ is an interesting function which I was able to
study manually (as much as I could) and now I want to plot it.

Answer 1

How? Use mathematical software such as Mathematica.

If all you wanted was a plot, I hope this answer will suffice. I tried doing this on Wolfram-Alpha but I think it's not possible as a free member.

Answer 2

As the comments and the Mathematica graph have established, the function $f$ has a positive branch and a negative branch.

On each branch, the function can be rewritten as $f(x,y)=g(x+y).$ That’s why you find all four second partial derivatives to be the same: the operators $\partial/\partial x$ and $\partial/\partial y$ act the same on $f$ and all its derivatives.

For very large $x+y$, on the negative branch we have a large negative value of $z$ so that $e^z\approx0$ and $z\approx-(x+y).$ But on the positive branch, since $e^z$ grows much faster than $z,$ the asymptotically are essentially $x+y$ versus $e^z$, so I believe we end up with $z\in \Theta(\log(x+y)).$ You can easily see that the first derivative you found matches those asymptotics on either branch.

The two branches meet along the line $x+y=1$, where the tangent plane is perpendicular to the $x,y$ plane, so the derivative is undefined there.