Here I tried the coordinate geometry, as in the equation represents a sphere. From there $(z-x)$ would imply the distance between the $z$ coordinate and $x$ coordinate so that the difference is maximum.
Since it is a sphere with radius $1$, $z-x =$ distance between $(0,0,1)$ and $(1,0,0)$; which would be $\sqrt 2$. Could there be any more approaches, calculus approach? (Since three variables are there thought it would be more complicated?)
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If $x^2+y^2+z^2=1$ then (using the inequality between geometric and arithmetic mean) $$ \tag{$*$} (z-x)^2 = x^2 - 2xz + z^2 \le 2(x^2+z^2) \le 2 $$ which implies $$ -\sqrt 2 \le z-x \le \sqrt 2 \, . $$ Equality holds in $(*)$ if $y=0$ and $x=-z$, so that the maximum of $z-x$ is attained at $(x, y, z) = (-1/\sqrt 2, 0, 1/\sqrt 2)$ .
If you'd like a calculus method, this situation is perfect for Lagrange Multipliers. Let $f(x,y,z)=z-x$ and $g(x,y,z) = x^2+y^2+z^2$. Since $\{x,y,z\in\mathbb{R}^n :g(x,y,z) = 1\}$ is a compact set, so by the Extreme Value Theorem, we know it has a maximum. Now, to find the possibilities for the maximum, we solve for $x,y,z,\lambda$ solving $g(x,y,z) = 1$ and $$\nabla f = \lambda \nabla g \implies \begin{pmatrix} -1\\0\\1 \end{pmatrix} = \lambda\begin{pmatrix} 2x,2y,2z \end{pmatrix}$$ Thus, $\lambda \neq 0$, so we can divide by it to get $y=0$ and $$\begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} -1/(2\lambda)\\ 0\\ 1/(2\lambda) \end{pmatrix}$$ And so solving for $x^2+y^2+z^2= 1$, we get $2\cdot 1/(4\lambda^2) = 1$ which implies $\lambda = \pm1/\sqrt{2}$. This gives the following two possible maxima: $$(x,y,z) = \left(\pm\frac{1}{\sqrt{2}}, 0, \mp \frac{1}{\sqrt{2}} \right)$$ which give $f(x) = \pm \sqrt{2}$ so the maximum is $\sqrt{2}$ (and the minimum is $-\sqrt{2}$).