Find the extrema of $f$ subject to the stated constraint.
\[
f(x, y, z)=x-y+z, \text { subject to } x^{2}+y^{2}+z^{2}=1
\]
maximum $(x, y, z)=$
minimum $\quad(x, y, z)=$

Step 1 ：The problem is asking us to find the maximum and minimum values of the function \(f(x, y, z) = x - y + z\) subject to the constraint \(x^2 + y^2 + z^2 = 1\). This is a problem of optimization with constraints, which can be solved using the method of Lagrange multipliers.

Step 2 ：We introduce a new variable, \(\lambda\), called the Lagrange multiplier, and consider the function \(L(x, y, z, \lambda) = f(x, y, z) - \lambda (g(x, y, z) - c)\), where \(g(x, y, z) = x^2 + y^2 + z^2\) and \(c = 1\). So, \(L(x, y, z, \lambda) = x - y + z - \lambda (x^2 + y^2 + z^2 - 1)\).

Step 3 ：We then find the critical points of \(L\) by setting its partial derivatives equal to zero. This gives us the system of equations: \(\frac{\partial L}{\partial x} = 1 - 2\lambda x = 0\), \(\frac{\partial L}{\partial y} = -1 - 2\lambda y = 0\), \(\frac{\partial L}{\partial z} = 1 - 2\lambda z = 0\), and \(\frac{\partial L}{\partial \lambda} = x^2 + y^2 + z^2 - 1 = 0\).

Step 4 ：Solving the first three equations for \(x\), \(y\), and \(z\) respectively gives \(x = \frac{1}{2\lambda}\), \(y = -\frac{1}{2\lambda}\), and \(z = \frac{1}{2\lambda}\). Substituting these into the fourth equation gives \(\left(\frac{1}{2\lambda}\right)^2 + \left(-\frac{1}{2\lambda}\right)^2 + \left(\frac{1}{2\lambda}\right)^2 - 1 = 0\), which simplifies to \(\frac{3}{4\lambda^2} - 1 = 0\).

Step 5 ：Solving this equation for \(\lambda\) gives \(\lambda = \pm \frac{\sqrt{3}}{2}\). Substituting these values back into the expressions for \(x\), \(y\), and \(z\) gives two possible solutions: \(\left(\frac{2}{\sqrt{3}}, -\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right)\) and \(\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}, -\frac{2}{\sqrt{3}}\right)\).

Step 6 ：Substituting these solutions into the function \(f(x, y, z)\) gives \(f\left(\frac{2}{\sqrt{3}}, -\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right) = \sqrt{3}\) and \(f\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}, -\frac{2}{\sqrt{3}}\right) = -\sqrt{3}\).

Step 7 ：Therefore, the maximum value of \(f(x, y, z)\) subject to the constraint \(x^2 + y^2 + z^2 = 1\) is \(\sqrt{3}\) and occurs at \(\left(\frac{2}{\sqrt{3}}, -\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right)\), and the minimum value is \(-\sqrt{3}\) and occurs at \(\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}, -\frac{2}{\sqrt{3}}\right)\).

Step 8 ：So, the maximum \((x, y, z) = \boxed{\left(\frac{2}{\sqrt{3}}, -\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right)}\) and the minimum \((x, y, z) = \boxed{\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}, -\frac{2}{\sqrt{3}}\right)}\).