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I want to show that the sphere with equation $x^2+y^2+z^2=1$ and the cone with equation $x^2+y^2=z^2$ are orthogonal at their intersection points.

What I did was let $F(x,y,z)=x^2+y^2+z^2$ and $G(x,y,z)=x^2+y^2-z^2$ so that $F=1$ and $G=0$ are level sets of some function of 3 variables.

Then I did $\nabla F \cdot \nabla G$ which gave me $4(x^2+y^2-z^2)=0$ since $G(x,y,z)=0.$

Is this correct have I done it legitimately?

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Yes, what you did is valid, although the last statement (since $G(x,y,z)=0$) is unnecessary and a bit misleading.

The point is that whenever $\nabla F\cdot \nabla G=0$ at some point which is on both $F= C_1$ and $G=C_2$, then the surfaces are orthogonal at that point, and if does not matter that you can or can't write $ \nabla F\cdot \nabla G = \kappa G$ and use $G=0$.

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    Yeah all I meant was hat $\nabla F \cdot \nabla G =4G$ and so $G=0$ means that dot product is zero hence they are perpendicular. – 2017-02-13