Need help completing the solution of an implicit function problem 
Let $f:\mathbb{R^3\rightarrow R^2}$ be given by $f(x,y,z)=(\cos(x)+5y-z^2-2, \exp(x)+2y-3z+3)$. Investigate whether the there exists a solution for $f(x,y,z)=(0,0)$ in a neighbourhood of $(0,1,2)$ find the values of $g(0)$ and $g'(0)$ such that $f(x,g(x))=0$ holds true.

By the implicit function theorem we can split the jacobian Matrix of $f$ as $D_1f$ and $D_2f$ since we treat the coordinates of $\mathbb{R^3}$ as $\mathbb{R^1 \times R^2}$ so we have that $$\begin{pmatrix}[D_1f|D_2f] \end{pmatrix}= \begin{pmatrix}-\sin(x)&5&-2z \\ e^x & 2&-3 \end{pmatrix} $$
Following the theorem again if the 2x2 matrix on the right hand side is invertible then there exists an open subset $U\in \mathbb{R}$ and point $a\in U$ so that function $g:U\rightarrow \mathbb{R^2}$ is continuosly differentiable and $f(x,g(x))=0$ holds true for all x in U. Since we are meant to investigate point $(x=0,y=1,z=2)$ then the matrix $D_2f$ has a determinant that's non zero and is therefore invertible so the function g exists. Furthermore the theorem tells us that: $$\begin{pmatrix} \frac{\partial g}{\partial x}\end {pmatrix}_{2\times1}=-(D_2f)^{-1}(D_1f)= \begin{pmatrix}1/7\big(3\sin(x)+4e^x\big) \\1/7\big(2\sin(x)+5e^x \big) \end {pmatrix}$$
Unless there are mistakes in my algebra $g'(0)=(4/7,5/7)$ but what I'm having trouble with is the value of $g(0)$. I don't recall the theorem having a particular method that helps find $g$ and when I tried to find an anti derivative to the expressions in 2x1 matrix the result of $g(0)$ was't equal to $(1,2)$. I'd appreciate if somone could clarify where I went wrong or if I'm missing something.

I followed your steps Alp. I would integrate $g_x$ you found immediately to find
$$g(x)=\left[\begin{array} .-\frac{3}{7}\cos x + \frac{4}{7}e^x \\ -\frac{2}{7}\cos x + \frac{5}{7}e^x\end{array}\right]+\left[\begin{array} .c_1 \\ c_2 \end{array}\right].$$
Now, $g(0)=\left[\begin{array} .1 \\ 2 \end{array}\right]$, so $c_1=\frac{6}{7}$ and $c_1=\frac{11}{7}$. We found the $g(x)$ wanted. I checked on the second component that $f(x,g(x))=0$.
