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) $.