Call $(E_k)$ the equation $y_k=\frac27y_{k+1}+\frac27y_{k-1}+\frac37y_{k-2}$. You are asked to find a bounded solution $(y_k)_{k\geqslant-1}$ of the system $(E_k)_{k\geqslant1}$ with the initial condition $y_{-1}=\frac13$, $y_0=1$. Nowhere does this involve a condition on $y_{-2}$.
A priori, the system $(E_k)_{k\geqslant1}$ has a unique solution $(y_k)_{k\geqslant-1}$ for every initial condition $(y_{-1},y_0,y_1)$ hence the missing initial condition $y_1$ is specified by the additional condition that $(y_k)_{k\geqslant-1}$ must stay bounded. Here is how.
As every linear system, this one can be reformulated as $Y_{k+1}=AY_k$ for every $k\geqslant1$, where $Y_k=(y_k,y_{k-1},y_{k-2})^T$, for some $3\times3$ matrix $A$, with the initial condition $Y_1=(y_1,1,\frac13)$. The eigenvalues of $A$ are $1$, $3$, and $a=-\frac12$, hence the fact that $(y_k)_{k\geqslant-1}$ stays bounded is equivalent to the fact that $Y_1$ has no component on the eigenvector of the eigenvalue $3$. Writing $Y_1$ as a linear combination of the eigenvectors $U=(1,1,1)^T$ for the eigenvalue $1$ and $V=(1,-2,4)^T$ for the eigenvalue $a$ imposes the condition $y_1=\frac23$.
Once $y_1$ is known, $Y_1$ is known as $Y_1=(\frac23,1,\frac13)^T=\frac79U-\frac19V$ and $Y_k=A^{k-1}Y_1$ hence $y_k=(1,0,0)A^{k-1}Y_1=(1,0,0)(\frac79U-\frac19a^{k-1}V)=\frac79-\frac19a^{k-1}=\frac79+\frac29a^{k}$ $(*)$.
Naturally, $(*)$ provides the values of $y_0$ and $y_1$ we started from, namely, $y_{-1}=\frac13$ and $y_0=1$, since these were part of our initial conditions. And $(*)$ also provides the unique value of $y_1$ which makes $(y_k)_{k\geqslant-1}$ stay bounded for these $y_{-1}$ and $y_0$, namely, $y_1=\frac23$. But none of this uses $y_{-2}$ and one could very well have indicated that $y_{-2}=\frac1{\sqrt6}$ or any other value without modifying the rest.