Suppose that $y_1(t), \ldots, y_n(t)$ are solutions of $\frac{d^n y}{dt} + p_{n-1}(t) \frac{d^{n-1} y}{dt} + \cdots + p_1(t) \frac{dy}{dt} + p_0(t) y = 0$, and suppose that their Wronskian is zero for $t = t_0$, i.e.
\begin{equation*}
\left|
\begin{array}{cccc}
y_1(t_0) & y_2(t_0) & \cdots & y_n(t_0) \\
y_1'(t_0) & y_2'(t_0) & \cdots & y_n'(t_0) \\
\vdots & \vdots & \ddots & \vdots \\
y_1^{(n-1)}(t_0) & y_2^{(n-1)}(t_0) & \cdots & y_n^{(n-1)}(t_0)
\end{array}
\right| = 0.
\end{equation*}
Then the corresponding matrix is not invertible, and the system of equations
\begin{array}{c}
c_1 y_1(t_0) &+& c_2 y_2(t_0) &+& \cdots &+& c_n y_n(t_0) &=& 0 \\
c_1 y_1'(t_0) &+& c_2 y_2'(t_0) &+& \cdots &+& c_n y_n'(t_0) &=& 0 \\
\vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\
c_1 y_1^{(n-1)}(t_0) &+& c_2 y_2^{(n-1)}(t_0) &+& \cdots &+& c_n y_n^{(n-1)}(t_0) &=& 0 \\
\end{array}
has a nontrivial solution for $c_1, c_2, \ldots, c_n$ not all zero.
Let $y(t) = c_1 y_1(t) + \cdots + c_n y_n(t)$. Because $y(t)$ is a linear combination of solutions of the differential equation, $y(t)$ is also a solution of the differential equation. Additionally, because the weights satisfy the above system of equations, we have $y(t_0) = y'(t_0) = \cdots = y^{(n-1)}(t_0) = 0$.
These initial conditions and the original differential equation define an initial-value problem, of which $y(t)$ is a solution. If $p_0(t), p_1(t), \ldots, p_{n-1}(t)$ are continuous, then any initial-value problem associated with the differential equation has a unique solution. Obviously $y^*(t) = 0$ is a solution of the initial-value problem; since we know that $y(t)$ is also a solution of the same initial-value problem, it follows that $y(t) = 0$ for all $t$, not just $t = t_0$.
We now have $c_1 y_1(t) + \cdots + c_n y_n(t) = 0$ for all $t$, where $c_1, \ldots, c_n$ are not all zero. Thus the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent.
Conversely, if the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent, then the system of equations
\begin{array}{c}
c_1 y_1(t) &+& c_2 y_2(t) &+& \cdots &+& c_n y_n(t) &=& 0 \\
c_1 y_1'(t) &+& c_2 y_2'(t) &+& \cdots &+& c_n y_n'(t) &=& 0 \\
\vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\
c_1 y_1^{(n-1)}(t) &+& c_2 y_2^{(n-1)}(t) &+& \cdots &+& c_n y_n^{(n-1)}(t) &=& 0 \\
\end{array}
has a nontrivial solution for every $t$, the corresponding matrix is not invertible for any $t$, and $W[y_1, \ldots, y_n](t) = 0$.
You can directly use Abel's identity to show that if the Wronskian of any two solutions of the differential equation $y''+p(x)y'+q(x)y = 0$ (on an interval $I$) is constant, then $p(x) = 0$. (If $\int_{x_0}^{x} p(t) dt$ is constant $\forall$ $x \in I$, then $p(t) = 0 $)
The answer to your last question can be found here.
Best Answer
The Wronskian of $y_1$ and $y_2$ is defined as
$W(y_1, y_2) = \det \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} = y_1y_2' - y_2 y_1'; \tag 1$
we may easily find the derivative
$W' = y_1'y_2' + y_1y_2'' - y_2'y_1' - y_2y_1'' = y_1y_2'' - y_2y_1''; \tag 2$
further progress is made using the given equation
$3y''+(6/x)y'+3e^xy = 0, \tag 3$
out of which the constant factor $3$ may be divided, leaving
$y'' + (2/x)y'+ e^xy = 0, \tag 4$
which we know $y_1$ and $y_2$ solve; thus we have
$W'$ $= y_1(-((2/x)y_2' + e^xy_2)) - y_2(-((2/x)y_1' + e^xy_1)) = -(2/x)y_1y_2' - e^xy_1y_2 + (2/x)y_2y_1' + e^xy_1y_2$ $= -(2/x)y_1y_2'+ (2/x)y_2y_1' = -(2/x)(y_1y_2' - y_2y_1') =-(2/x)W; \tag 5$
once the clutter of this equation is removed we are left with
$W' = -(2/x)W, \tag 6$
which is a case of Abel's identity; the unique solution to (6) taking the value $W(1)$ at $1$ is
$W(x) = W(1)\exp \left (-\displaystyle \int_1^x (2/s)\;ds \right); \tag 7$
we may easily evaluate the integral:
$\displaystyle \int_1^x (2/s)\;ds = 2\int_1^x (1/s)\;ds = 2(\ln x - \ln 1) = 2\ln x = \ln x^2; \tag 8$
thus,
$W(x) = W(1)\exp(-\ln x^2) = W(1)\exp(\ln x^{-2}) = W(1)x^{-2}. \tag 9$
Now with
$W(1) = 2, \; x = 10, \tag{10}$
$W(x) = 2(10)^{-2} = 2(.01) = .02. \tag{11}$