Finding $x^2+y^2+z^2$ given that $x+y+z=0$, $x^3+y^3+z^3=3$ and $x^4+y^4+z^4=15$

Yes, this can be solved without guessing, using Newton's identities.

Since $x + y + z = 0$, they are the roots of $t^3 + at -b = 0$.

Newton's identities give us (in a straightforward mechanical manner) that

$$x^3 + y^3 + z^3 = 3b$$ $$x^4 + y^4 + z^4 = 2a^2$$

and $$x^2 + y^2 + z^2 = -2a$$

This gives us $b=1$ and $2a^2 = 15$.

You can solve for $a$ and find the value of $x^2 + y^2 + z^2 = -2a$. Note that if you assume $x,y,z$ are real, then you need to pick $a \lt 0$ which gives us

$$x^2 + y^2 + z^2 = \sqrt{30}$$

Note that you did not even need to use the value of $x^3 + y^3 + z^3$.

You can also derive the above identities in a slightly tedious way, as in my answer here: Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$

Note that $0 = (x+y+z)^2 = (x^2+y^2+z^2) + 2(xy+xz+yz)$, so $x^2+y^2+z^2 = -2(xy+xz+yz)$.

And $$\begin{align*} 15 &= x^4+y^4+z^4\\ & = (x^2+y^2+z^2)^2 - 2(x^2y^2 + x^2z^2 + y^2z^2).\end{align*}$$

And $$\begin{align*} (xy+xz+yz)^2 &= (x^2y^2 + x^2z^2 + y^2z^2) + 2(x^2yz+xy^2z+xyz^2)\\ &= (x^2y^2+x^2z^2 + y^2z^2) + 2xyz(x+y+z)\\ &= x^2y^2+x^2z^2+y^2z^2 + 2xyz(0)\\ &= x^2y^2+x^2z^2+y^2z^2 \end{align*}$$ hence $$x^2y^2+x^2z^2+y^2z^2 = (xy+xz+yz)^2.$$ So $$\begin{align*} 15 &= x^4+y^4+z^4\\ &= (x^2+y^2+z^2)^2 - 2(x^2y^2 + x^2z^2+y^2z^2)\\ &= (x^2+y^2+z^2)^2 - 2(xy+xz+yz)^2\\ &= (x^2+y^2+z^2)^2 -2(xy+xz+yz)(xy+xz+yz)\\ &= (x^2+y^2+z^2)^2 +(x^2+y^2+z^2)(xy+xz+yz)\\ &= (x^2+y^2+z^2)^2 + (x^2+y^2+z^2)\left(-\frac{1}{2}(x^2+y^2+z^2)\right)\\ &= (x^2+y^2+z^2)^2 -\frac{1}{2}(x^2+y^2+z^2)^2\\ &=\frac{1}{2}(x^2+y^2+z^2)^2 \end{align*}$$ so $(x^2+y^2+z^2)^2 = 30$, hence $x^2+y^2+z^2=\sqrt{30}$.