Solution Verification: Differentiate $\left(x-1/2\right)^2+\left(y-1/4\right)^2=5/16$ with respect to $x$ Differentiate $$\:\left(x-1/2\right)^2+\left(y-1/4\right)^2=5/16$$

With respect to $x.$

The above simplifies to

$$x^2-x+y^2-\frac{y}{2} = 0$$

I know that you have to take the implicit derivative, such that

$$\frac{d}{dx}(x^2-x+y^2-\frac{y}{2}) = \frac{d}{dx}(0)$$

Thus

$$2x-1+2y-\frac{1}{2}=0$$

and

$$y=-x+\frac{3}{4}$$

So $$y'=-x+\frac{3}{4}$$ 

Have I made any errors, and where can I improve on this solution? In particular, I am worried because the last two equations are the same except the difference between $y$ and $y'.$ I mean, both can't be assumed to be true, right?

And I am asked to find the equation of the tangent of the circle that intersects at the point $(1, 0).$ In this case, do I write the equation in the form $y=mx+b$ (but with numbers replacing $m$ and $b$) as well, in which case I would have two $y$ variables in my solution?
Your error is first forgetting that $y$ depends on $x$ (I assume because implicit derivatives are mentioned and $x(y)$ would be pretty mean) and then getting confused and solving for $y$ and not $y'$. 

You are right that you need to take 
$$
\frac{d}{dx}(x^2-x+y^2-y/2)=0
$$
But $y$ is a function of $x$ so this should be by the chain rule
$$
2x-1+2yy'-1/2y'=0
$$
Can you solve for $y'$ now?

Now as for the tangent line to the circle described by this curve, you now have the equation for the slope of the tangent line to the curve at any point and a given point. So your $m=y'(1)$ and you can use precalculus methods to find the equation of the line in point slope form.
It should be $2x-1+\left(2y-\frac{1}{2}\right)y'=0$