Answer
$y=\left(\frac{\sec{x}+\tan{x}+C}{\tan{x}+\sec{x}}\right)^{-\frac{1+\sqrt{3}}{2}}$
Work Step by Step
Given
$$(1-\sqrt{3})y'+y\sec{x}=y^{\sqrt{3}}\sec{x}$$
Note the equation is in Bernoulli form: $\frac{dy}{dx}+P(x)y=Q(x)y^n$
Now divide all terms by $y^n$:
$$(1-\sqrt{3})y^{-\sqrt{3}}y'+y^{1-\sqrt{3}}\sec{x}=\sec{x}$$
Let $u=y^{1-\sqrt{3}}$. Thus, $\frac{1}{1-\sqrt{3}}\frac{du}{dx}=y^{-\sqrt{3}}\frac{dy}{dx}$
$$(1-\sqrt{3})\left(\frac{1}{1-\sqrt{3}}\right)\frac{du}{dx}+u\sec{x}=\sec{x}$$
$$\frac{du}{dx}+u\sec{x}=\sec{x}$$
Now solve as a first-order linear differential equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$
First, find the integration factor $I(x)=e^{\int{P(x)}dx}$:
$$I(x)=e^{\int{\sec{x}}dx}=e^{\ln{|\tan{x}+\sec{x}|}}=\tan{x}+\sec{x}$$
Multiply both sides by $I(x)$ and replace the left side with $\frac{d}{dx}[I(x)u]$:
$$\frac{d}{dx}[(\tan{x}+\sec{x})u]=\sec{x}(\tan{x}+\sec{x})$$
$$\frac{d}{dx}[(\tan{x}+\sec{x})u]=\sec{x}\tan{x}+\sec^2{x}$$
Integrate both sides:
$$\int{\frac{d}{dx}[(\tan{x}+\sec{x})u]}=\int{(\sec{x}\tan{x}+\sec^2{x})}dx$$
$$(\tan{x}+\sec{x})u=\sec{x}+\tan{x}+C$$
Solve for $y$:
$$(\tan{x}+\sec{x})y^{1-\sqrt{3}}=\sec{x}+\tan{x}+C$$
$$y^{1-\sqrt{3}}=\frac{\sec{x}+\tan{x}+C}{\tan{x}+\sec{x}}$$
$$y=\left(\frac{\sec{x}+\tan{x}+C}{\tan{x}+\sec{x}}\right)^{-\frac{1+\sqrt{3}}{2}}$$
