Your system
$$\left\{
\begin{array}{c}
\frac{1}{300}a-\frac{1}{200}b=5 \\
-\frac{1}{300}a+\left( -\frac{1}{300}+\frac{1}{200}\right) b-\frac{1}{200}
c=-e^{b} \\
-\frac{1}{200}b+\frac{1}{200}c=-e^{c}\tag{1}
\end{array}
\right. $$
is equivalent to
$$\left\{
\begin{array}{c}
2a-3b=3000 \\
-2a+b-3c=-600e^{b} \\
-b+c=-200e^{c}\tag{2}
\end{array}
\right. $$
and to
$$\left\{
\begin{array}{c}
a=\frac{3}{2}\left( c+200e^{c}\right) +1500 \\
-5c-400e^{c}-3000+600\text{ exp}(c+200e^{c})=0 \\ \tag{3}
b=c+200e^{c}.
\end{array}
\right. $$
The second equation has two solutions (computation in WolframAlpha):
$c\approx -600.000$ and $c\approx -3.64058$.
The Newton's method applied to $$f(c)=-5c-400e^{c}-3000+600\text{ exp}(c+200e^{c})\tag{4}$$
consists of the following iterations
$$c_{k+1}=c_{k}-\frac{f(c_{k})}{f^{\prime }(c_{k})},\qquad k=1,2,\dots\tag{5}$$
with
$$f'(c)=-5-400e^{c}+600\left( 1+200e^{c}\right) \text{ exp}(c+200e^{c})\tag{6}$$
Starting with e.g. $c_{1}=-500$, we get
$$
\begin{eqnarray*}
f(c_{1}) &=&-5c_{1}-400e^{c_{1}}-3000+600\text{ exp}(c_{1}+200e^{c_{1}}) \\
f(-500) &=&5\times 500-400e^{-500}-3000+600\text{ exp}(-500+200e^{-500}) \\
&\approx &-500.0
\end{eqnarray*}$$
and
$$
\begin{eqnarray*}
f^{\prime }(c_{1}) &=&-5-400e^{c_{1}}+600\left( 1+200e^{c_{1}}\right)
\text{ exp}(c_{1}+200e^{c_{1}}) \\
f^{\prime }(-500) &=&-5-400e^{-500}+600\left( 1+200e^{-500}\right)
\text{ exp}(-500+200e^{-500}) \\
&\approx &-5.0.
\end{eqnarray*}$$
And so,
$$
c_{2}=c_{1}-\frac{f(c_{1})}{f^{\prime }(c_{1})}\approx -500-\frac{-500.0}{
-5.0}\approx -600.0,
$$
which is already a good approximation.
For $c_{1}=-3.6$, we get successively
$$\begin{eqnarray*}
c_{2} &\approx &-3.6-\frac{879.64}{25019.0}\approx -3.6352 \\
c_{3} &\approx &-3.6352-\frac{71.578}{19406.}\approx -3.6389 \\
c_{4} &\approx &-3.6389-\frac{1.\,417\,3}{18902.}\approx -3.6390 \\
c_{5} &\approx &-3.6390-\frac{29.\,615}{18889.}\approx -3.6406 \\
c_{6} &\approx &-3.6406-\frac{-0.435\,93}{18676}\approx
-3.6406.
\end{eqnarray*}$$
From $(3)$ for $c\approx -3.6406$, we get the solution $(a,b,c)\approx
(1502.4,1.6067,-3.6406)$ and for $c\approx -600.0$ the solution $(a,b,c)\approx
(600.0,-600.0,-600.0)$.

Plot of $f(c)$ for $c=-700$ to $c=-3.62$

Plot of $f(c)$ for $c=-3.7$ to $c=-3.6$
Added: You have perhaps in mind the general method for solving a nonlinear
system, applied to the present case. Let's denote $x_{1}=a,x_{2}=b,x_{3}=c$. The system
$$
\begin{pmatrix}
f_{1}(x_{1},x_{2},x_{3}) \\
f_{2}(x_{1},x_{2},x_{3}) \\
f_{3}(x_{1},x_{2},x_{3})
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0
\end{pmatrix},\tag{7}$$
where
$$
\begin{pmatrix}
f_{1}(x_{1},x_{2},x_{3}) \\
f_{2}(x_{1},x_{2},x_{3}) \\
f_{3}(x_{1},x_{2},x_{3})
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{300}x_{1}-\frac{1}{200}x_{2}-5 \\
-\frac{1}{300}x_{1}+\left( -\frac{1}{300}+\frac{1}{200}\right) x_{2}-\frac{1
}{200}x_{3}+e^{x_{2}} \\
-\frac{1}{200}x_{2}+\frac{1}{200}x_{3}+e^{x_{3}}
\end{pmatrix},\tag{8}$$
has the Jacobian matrix
$$\begin{eqnarray*}
J\left( x\right) &=&
\begin{pmatrix}
\frac{\partial f_{1}(x_{1},x_{2},x_{3})}{\partial x_{1}} & \frac{\partial
f_{1}(x_{1},x_{2},x_{3})}{\partial x_{2}} & \frac{\partial
f_{1}(x_{1},x_{2},x_{3})}{\partial x_{3}} \\
\frac{\partial f_{2}(x_{1},x_{2},x_{3})}{\partial x_{1}} & \frac{\partial
f_{2}(x_{1},x_{2},x_{3})}{\partial x_{2}} & \frac{\partial
f_{2}(x_{1},x_{2},x_{3})}{\partial x_{3}} \\
\frac{\partial f_{3}(x_{1},x_{2},x_{3})}{\partial x_{1}} & \frac{\partial
f_{3}(x_{1},x_{2},x_{3})}{\partial x_{2}} & \frac{\partial
f_{3}(x_{1},x_{2},x_{3})}{\partial x_{3}}
\end{pmatrix}
\\
&=&
\begin{pmatrix}
\frac{1}{300} & -\frac{1}{200} & 0 \\
-\frac{1}{300} & \frac{1}{600}+e^{x_{2}} & -\frac{1}{200} \\
0 & -\frac{1}{200} & \frac{1}{200}+e^{x_{3}}
\end{pmatrix}
.
\end{eqnarray*}\tag{9}
$$
The Newton's method consists of starting with an approximation $x^{(1)}$ and
find successively
$$x^{(k+1)}=x^{(k)}+\Delta x^{(k)},\qquad k=1,2,\dots\tag{10}$$
where $\Delta x^{(k)}$ is a solution of
$$J\left( x^{(k)}\right) \Delta x^{(k)}=-f\left( x^{(k)}\right) ,\tag{11}$$
which can be found by Gaussian elimination.
Notation: $x^{(k)}$ is the vector $\left(
x_{1}^{(k)},x_{2}^{(k)},x_{3}^{(k)}\right) ^{T}$, $x^{(k+1)}$ is the vector $%
\left( x_{1}^{(k+1)},x_{2}^{(k+1)},x_{3}^{(k+1)}\right) ^{T}$, $J\left(
x^{(k)}\right) $ is the Jacobian matrix evaluated at $\left(
x_{1}^{(k)},x_{2}^{(k)},x_{3}^{(k)}\right) $ and $f\left( x^{(k)}\right)$ is the vector column $(8)$ of $f$ evaluated at $\left(
x_{1}^{(k)},x_{2}^{(k)},x_{3}^{(k)}\right) $.