At $f\left(x\right)={\sin }^{6}x+{\cos }^{6}x+k({\sin }^{4}x+{\cos }^{4}x)$ . If $f\left(x\right)=0$  has a solution then set of values of $k$  are

$f\left(x\right)=0$ 

$\Rightarrow (1-3{\sin }^{2}x{\cos }^{2}x)$ $+k[1-2{\sin }^{2}x{\cos }^{2}x]=0$

$\Rightarrow k=\frac{3{\sin }^{2}x{\cos }^{2}x-1}{1-2{\sin }^{2}x{\cos }^{2}x}$ $=-\frac{3}{2}\frac{1-2{\sin }^{2}x{\cos }^{2}x-\frac{1}{3}}{1-2{\sin }^{2}x{\cos }^{2}x}$

$=-\frac{3}{2}\left(1-\frac{\frac{1}{3}}{1-2{\sin }^{2}x{\cos }^{2}x}\right)$

Minimum of ${\sin }^2\theta {\cos }^2\theta =0$  at $\theta =0,\pi /2$

$\frac{-3}{2}\left(1-\frac{{\displaystyle \frac{1}{3}}}{1-0}\right)=\frac{-3}{2}\left(\frac{2}{3}\right)=-1$

Maximum of ${\sin }^2\theta {\cos }^2\theta =1/4$  at $\theta =\pi /4$   

$\frac{-3}{2}\left(1-\frac{{\displaystyle \frac{1}{3}}}{1-2\left({\displaystyle \frac{1}{4}}\right)}\right)=\frac{-3}{2}\left(1-\frac{2}{3}\right)=-\frac{1}{2}$

  Hence $k\in \left[-1,-\frac{1}{2}\right]$