How can i show that $0$ is a regular value of $F:\mathbb{R}^{3} \rightarrow \mathbb{R}^2$ given by $$F(x,y,z)=(x^3+y^3+z^3-1,z-xy)$$ I compute the diferential of $F$ obtaining: $$dF=\pmatrix{ 3x^2 & -y\cr 3y^2 & -x\cr 3z^2 &1}$$ and i know that if $det[(dF)(dF)^t] \neq 0$ then $F$ is a submersion but compute this determinant and see the points where it is $0$ its hard.
Any hint or alternatively way to show that i will be very grateful.
To solve this problem, you have to play with the equations $x^3+y^3+z^3 = 1$ and $z=xy$.
Let $Z = F^{-1}(0)$. Let $(x,y,z)\in Z$, and consider the matrix of $dF_{(x,y,z)}$ in the canonical basis, which is $$ \mathrm{Mat}_{\mathrm{can}}(dF_{(x,y,z)}) = \begin{pmatrix}3x^2 & 3y^2 & 3z^2 \\ -y & -x & 1 \end{pmatrix} = \begin{pmatrix}3x^2 & 3y^2 & 3x^2y^2 \\ -y & -x & 1 \end{pmatrix}, $$ since on $Z$, $z=xy$. (Note that this is the transpose of what you have stated: indeed, $dF_{(x,y,z)}$ goes from $\Bbb R^3$ to $\Bbb R^2$, not the converse.)
Let us show that this matrix has rank $2$. First of all, $x$ and $y$ cannot be simultaneously vanishing: By contradiction, if $x=y=0$, then $x^3+y^3+z^3-1=0$ gives $z=1$, while $z-xy=0$ gives $z=0$.
Without loss of generality, let us assume that $x\neq 0$. Consider the determinant of the matrix formed by the first and last columns: $$ \begin{vmatrix}3x^2 & 3x^2y^2 \\ -y & 1\end{vmatrix} = 3x^2 + 3x^2y^3 = 3x^2(1+y^3). $$ It vanishes if and only if $y^3=-1$, that is if and only if $y=-1$. Assume by contradiction that $y=-1$. Then $z-xy=0$ leads to $z=-x$. Hence, $x^3+y^3+z^3-1=0$ leads to $-2=0$, which is obviously false. If follows that $y\neq -1$ and that $dF_{(x,y,z)}$ has rank $2$. Hence, $(x,y,z)$ is a regular point of $F$.
This being true for all $(x,y,z)$ in $Z$, $0$ is a regular value for $F$.