#### Answer

See the explanation/proof below.

#### Work Step by Step

When the tangent line is horizontal, the slope of it at that point is zero. Therefore, we need to show that the slope of the tangent line is not equal to zero at any point. But, the slope of the tangent line is the derivative of the function at that point. Therefore, we need to prove that $\frac{dy}{dx}\ne0$ at any point.
Implicitly differentiating $x^{2}-3xy+y^{2}=1$, we get
$2x-3(1\times y+x\times \frac{dy}{dx})+2y\frac{dy}{dx}=0$
Or $ 2x-3y-3x\frac{dy}{dx}+2y\frac{dy}{dx}=0$
Or $-3x\frac{dy}{dx}+2y\frac{dy}{dx}=3y-2x$
Or $\frac{dy}{dx}=\frac{3y-2x}{-3x+2y}$
$\frac{dy}{dx}=0\implies \frac{3y-2x}{-3x+2y}=0$
$\implies 3y=2x$ or $y=\frac{2x}{3}$
The tangent line is horizontal if and only if $y=\frac{2x}{3}$.
Substituting $y=\frac{2x}{3}$ in $x^{2}-3xy+y^{2}=1$, we obtain
$x^{2}-3x(\frac{2x}{3})+(\frac{2x}{3})^{2}=1$
$\implies -x^{2}+\frac{4x^{2}}{9}=1$
$\implies \frac{-5x^{2}}{9}=1$
Or $x^{2}=-\frac{9}{5}$
The square of a number cannot be negative. So, we showed that there is no value of $x$ for which the tangent line is horizontal.