Lagrange multipliers... Three variables. Let $f:\mathbb{R}^3\mapsto \mathbb{R}$ be defined by
\begin{equation*}
f(x,y,z) := x+3y+5z.
\end{equation*}
Compute the minimum of $f$ subject to the constraint that
\begin{equation*}
g(x,y,z) = x^2+y^2+z^2-1 = 0.
\end{equation*}
The method of Lagrange multipliers is as follows: We first solve $\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$ and $g(x,y,z) = k$ then plug in all solutions, $(x,y,z)$, into $f(x,y,z)$ and identify the minimum and maximum values, provided they exist and $\nabla g\neq 0$ at the point. Let's apply this to our problem.
We have $1 = 2\lambda x$, $3 = 2\lambda y$ and $5 = 2\lambda z$ along with $x^2+y^2+z^2 = 1$. Solving for $x$, $y$ and $z$ we get
\begin{equation*}
x = \frac{1}{2\lambda}, \quad y = \frac{3}{2\lambda}, \quad z = \frac{5}{2\lambda}
\end{equation*}
and plugging into our constraint gives
\begin{equation*}
x^2+y^2+z^2 = \frac{1}{4\lambda^2}+\frac{9}{4\lambda^2}+\frac{25}{4\lambda^2} = 1 \Longleftrightarrow \lambda^2 = \frac{35}{4},
\end{equation*}
or $\lambda = \pm\frac{\sqrt{35}}{2}$. For $\lambda = \frac{\sqrt{35}}{2}$, $(x,y,z) = \left(\frac{1}{\sqrt{35}},\frac{3}{\sqrt{35}},\frac{5}{\sqrt{35}}\right)$ and for $\lambda = -\frac{\sqrt{35}}{2}$, $(x,y,z) = \left(-\frac{1}{\sqrt{35}},-\frac{3}{\sqrt{35}},-\frac{5}{\sqrt{35}}\right)$. Substituting these into $f(x,y,z)$ gives
\begin{equation*}
f\left(\frac{1}{\sqrt{35}},\frac{3}{\sqrt{35}},\frac{5}{\sqrt{35}}\right) = \frac{1}{\sqrt{35}}+\frac{9}{\sqrt{35}}+\frac{25}{\sqrt{35}} = \frac{35}{\sqrt{35}} = \sqrt{35}
\end{equation*}
which is the maximum value of $f$, and
\begin{equation*}
f\left(-\frac{1}{\sqrt{35}},-\frac{3}{\sqrt{35}},-\frac{5}{\sqrt{35}}\right) = -\frac{1}{\sqrt{35}}-\frac{9}{\sqrt{35}}-\frac{25}{\sqrt{35}} = -\frac{35}{\sqrt{35}} = -\sqrt{35}
\end{equation*}
which is the minimum value of $f$ as required.
Is this all good? Thanks!

Yes, your computations and your result are correct. Using the Cauchy-Schwarz inequality, we can also obtain the same result very quickly:
Note that (where $\langle\cdot,\cdot\rangle$ is the euclidean inner product and $\|\cdot\|$ is the euclidean norm) $$f(x,y,z)=\left\langle(x,y,z),(1,3,5)\right\rangle \text{ and that } g(x,y,z)=0 \text{ means that }\|(x,y,z)\|=1.$$
Hence, by Cauchy-Schwarz, $$\lvert f(x,y,z)\rvert\le\|(x,y,z)\|\cdot\|(1,3,5)\|=1\cdot\|(1,3,5)\|=\sqrt{35}.$$
Equality is achieved if and only if $(x,y,z)$ is a scalar multiple of $(1,3,5)$, which, together with $\|(x,y,z)\|=1$, implies that we have the two global minimizers/maximizers $$(x,y,z)=\pm\frac1{\sqrt{35}}(1,3,5)$$
