You've surely been trying to express one of the unknowns using the others and plugging in. Try to preserve the symmetry between them instead:
$$(x+y+z)^3 = x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3y^2x + 3y^2z + 3z^2x + 3z^2y + 6xyz$$
But we know that the left hand side is a zero and, moreover, wherever something like $x+z$ appears we can replace that by a $-y$ (again, strictly keeping exchange symmetry). So the same equation can be written as
$$0 = x^3 + y^3 + z^3 + 3x^2(-x) + 3y^2(-y) + 3z^2(-z) + 6xyz$$
and even better than that!
$$\begin{aligned}
0 &= -2(x^3 + y^3 + z^3) + 6xyz \\
xyz &= \frac{x^3 + y^3 + z^3}3.
\end{aligned}$$
Surely replacing the right hand side by the left in your problem will make the multiplications simpler! Try to proceed from here :-)
Hint: apart from $S_{111} = xyz$ and $S_k = x^k+y^k+z^k$, among which you have a first relation here, you'll need other symmetric forms like $S_{11} = xy+xz+yz$.
Update: I have been able to derive both your equations this way so it's guaranteed to work. It just takes some time. There's a whole theory about monomial and power-sum symmetric polynomials but I didn't want to go into that.
Starting from $S_1 = 0$, you can derive things like
$$\begin{aligned}
S_2 &= -2S_{11} \\
S_3 &= 3S_{111} \\
S_4 &= \frac12 S_2^2
\end{aligned}$$
and also some useful observations like
$$\begin{aligned}
S_{kl} &= S_k S_l - S_{k+l} \\
S_{(a+c)(b+c)c} &= S_{ccc} S_{ab} = S_{111}^c S_{ab}
\end{aligned}$$
With these you're one little theorem away from $S_5/5 = S_3S_2/6$ and $S_7/7 = S_5S_2/10$.
