Answer
$$y'=-\frac{y}{x\ln x}$$
Work Step by Step
$$x^y=\sqrt2$$
The presence of $x^y$ prevents us from applying immediately the methods of implicit differentiation. Instead, we need to take the natural logarithm of both sides first: $$\ln(x^y)=\ln\sqrt2$$
We have $\ln(x^y)=y\ln x$ $$y\ln x=\ln\sqrt2$$
Now using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$y'\ln x+y(\ln x)'=(\ln\sqrt2)'$$
$$y'\ln x+\frac{y}{x}=0$$
Next, we need to separate elements with $y'$ and those without $y'$ into 2 sides of the equation:
$$y'\ln x=-\frac{y}{x}$$
Finally, calculate for $y'$: $$y'=-\frac{y}{\frac{x}{\ln x}}=-\frac{y}{x\ln x}$$
