Prove that surrounded by zero exist smooth function $y(x)$ 
Let the function $G$ be such that $G(x)=\sum_{ij}G_{ij}(x)x_ix_j$ for some $G_{ij}\in C^{\infty}$ that vanish at zero.   Prove that in a neighborhood of zero there  exists a  smooth function $y(x)$ such that:
$$Q(x+y(x))=Q(x)+G(x), y(0)=0, dy(0)=0$$ where $Q(x)$ is a nondegenerate quadratic form.

I have an idea to use $F(x,y)=Q(x+y)-Q(x)-G(x)$ and then use the theorem:
Let $r$ - set of all smooth functions vanishing at zero. Assume that for smooth function $F$ we can write $F(x,0)$ as $F(x,0)=\sum_{ij}\frac{\partial F}{\partial y_i}(x,0) \frac{\partial F}{\partial y_j}(x,0)\phi_{ij}(x)$ for some $\phi_{ij}\in r$. Then $F(x,y(x))=0$ has in the surroundings $x=0$ smooth solution $y(x)$ such that $y_i(x)=\sum_{j} \frac{\partial F}{\partial y_j}(x,0)z_{ij}(x)$ for some $z_{ij}\in r$.
My idea stems from the fact that $F(x,0)=Q(x)-Q(x)-G(x)=-G(x)$ but I don't really know what to do with it.
Can I ask for help?

This is a way of deriving the OP's desired result, using a deus ex machina argument that I happened to know.
Given $G$ and a non-degenerate quadratic form $Q$ as stated, the
function $f(x)=Q(x)+G(x)$ has a non-degenerate critical point at $x=0$, with $f(0)=0$, $df(0)=0$, and $d^2f(0)=2Q$.  By the Morse Lemma there exists a diffeomorphism $u$ in a neighborhood of $0$ such that $f(x)= Q(u(x))$, with $u(0)=0$.  Let $\epsilon(x)=u(x)-(u(0)+du(0)x)=u(x)-du(0)x$ be the remainder of the linear Taylor approximation to $u$ at $0$.  We have $\epsilon(x)=o(\|x\|)$.
By matching Taylor expansions of $f(x)$ and $Q(u(x))=Q(0+du(0)x+o(\|x\|))$ through quadratic terms, we see that $Q(du(0)v)=Q(v)$ for all $v$.
That is, the coefficient of $x_ix_j$ in the Taylor expansion of $f(x)$ is the same as that of $Q(x)$, which is $2q_{ij}$.  The corresponding coefficient for $Q(u(x))$ is $2\sum_{kl} q_{kl}(\partial_i u_k)(\partial_j u_l)$.  Since $f(x)=Q(u(x))$ for all $x$ close to $0$, the coefficients of $x_ix_j$ must match.  
Thus, $f(x) = Q(u(x))=Q(du(0)^{-1}u(x))=Q(x+y(x))$, where $y(x)=du(0)^{-1}\epsilon(x)$.
This is the desired formula.  The matching Taylor expansion verifies that $du(0)\in \mathcal O(Q)$, in the notation of this post; this step is needed because of the notational mismatch between the way the current problem is stated and the more coordinate-free way the Morse Lemma is usually stated.
