The determinant
$$
\det(A) = \det(a_1, a_2, a_3)
$$
is an alternating multi linear form, this means
$$
\det(a_1+ \alpha a_2, a_2, a_3) = \det(a_1, a_2, a_3) + \alpha \det(a_2, a_2, a_3)
$$
because $\det$ is linear in the first argument as multi linear form.
Because it is an alternating form, if we exchange the first two arguments, the sign changes and we have
$$
\det(a_2, a_2, a_3) = - \det(a_2, a_2, a_3)
$$
which means $\det(a_2,a_2,a_3)$ vanishes.
This leads to
$$
\det(a_1+ \alpha a_2, a_2, a_3) = \det(a_1, a_2, a_3)
$$
and similar for the other arguments.
We can use this property to simplify your determinant:
\begin{align}
d &=
\begin{vmatrix}
y + z & x & x \\
y & z + x & y \\
z & z & x + y
\end{vmatrix}
=
\begin{vmatrix}
0 & -2z & -2y \\
y & z + x & y \\
z & z & x + y
\end{vmatrix}
=
2
\begin{vmatrix}
0 & -z & -y \\
y & z + x & y \\
z & z & x + y
\end{vmatrix}
\\
&=
2
\begin{vmatrix}
0 & -z & -y \\
y & x & 0 \\
z & 0 & x
\end{vmatrix}
=
-2
\begin{vmatrix}
0 & z & y \\
y & x & 0 \\
z & 0 & x
\end{vmatrix}
=
2
\begin{vmatrix}
0 & y & z \\
y & 0 & x \\
z & x & 0
\end{vmatrix}
\end{align}
If $x = y = z = 0$ then $d = 0$ and the proposed formula holds.
If two of the variables $x,y,z$ are zero, then one of the columns
is zero and we have e.g.
$$
\det(0, a_2, a_3) = \det(0 + a_2, a_2, a_3) = \det(a_2, a_2, a_3) = 0
$$
If e.g. $x = 0$ we have
$$
d =
2
\begin{vmatrix}
0 & y & z \\
y & 0 & 0 \\
z & 0 & 0
\end{vmatrix}
=
2 y z
\begin{vmatrix}
0 & 1 & 1 \\
y & 0 & 0 \\
z & 0 & 0
\end{vmatrix}
=
2 y z
\begin{vmatrix}
0 & 0 & 1 \\
y & 0 & 0 \\
z & 0 & 0
\end{vmatrix}
=
-2 y z
\begin{vmatrix}
0 & 0 & 1 \\
0 & y & 0 \\
0 & z & 0
\end{vmatrix}
= 0
$$
and similar if just $y$ or $z$ vanishes.
If $x,y,z$ do not vanish each, we have
\begin{align}
d
&=
2
\begin{vmatrix}
0 & y & z \\
y & 0 & x \\
z & x & 0
\end{vmatrix}
=
\frac{2}{xyz}
\begin{vmatrix}
0 & xy & xz \\
yz & 0 & xz \\
yz & xy & 0
\end{vmatrix}
=
\frac{2}{xyz}
\begin{vmatrix}
0 & xy & xz \\
yz & 0 & xz \\
yz & 0 & -xz
\end{vmatrix}
\\
&=
\frac{2}{xyz}
\begin{vmatrix}
0 & xy & xz \\
0 & 0 & 2xz \\
yz & 0 & -xz
\end{vmatrix}
=
\frac{4}{y}
\begin{vmatrix}
0 & xy & xz \\
0 & 0 & 1 \\
yz & 0 & -xz
\end{vmatrix}
=
\frac{4}{y}
\begin{vmatrix}
0 & xy & 0 \\
0 & 0 & 1 \\
yz & 0 & 0
\end{vmatrix}
\\
&=
4xyz
\begin{vmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{vmatrix}
=
-4xyz
\begin{vmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{vmatrix}
=
4xyz
\begin{vmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{vmatrix}
\\
&= 4xyz
\end{align}
Best Answer
You can find these facts in most introductory linear algebra texts, such Gilbert Strang "Linear Algebra and Its Applications" or David Poole's "Linear Algebra and its Applications" (I like the presentation in this book), and I'd recommend looking at these instead. There are some in "Linear Algebra Done Wrong", but I dislike this book.
Theres also the Wikipedia article.