The circles are $x^2+ y^2 = 1$ and $(x-3)^2 + y^2 = 4$
It is given in the question that there is a line with a positive slope that is tangent to both circles. To attempt to solve for this I rearranged both equations so that $y$ was a function of $x$ and only considered the positive halves of the circles. After I took the derivative of both and set them equal to each other I found solutions $x=1$ and $-3$. How is $x=-3$ a solution? It's not on either circle. What does it mean for $x = -3$ to be a solution to this equation?
Your solution technique is unlikely to work. The points of intersection of the line with each circle should not both have the same $x$ coordinate. (I'll say more about this at the end.)
But let's pretend that what we actually want to do is find one $x$ value (i.e., one vertical line on the Cartesian plane) where both circles have the same slope of the tangent (which is what you did). We calculate \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \sqrt{1-x^2} &= \frac{-x}{\sqrt{1-x^2}} \text{ and } \\ \frac{\mathrm{d}}{\mathrm{d}x} \sqrt{4-(x-3)^2} &= \frac{3-x}{\sqrt{4-(x-3)^2}} \end{align*} These are the slopes for the upper halves of the circles. We set these equal and solve. \begin{align*} \frac{-x}{\sqrt{1-x^2}} &= \frac{3-x}{\sqrt{4-(x-3)^2}} \end{align*} Amusingly, the only solution to this equation is $x = -3$. The step we are about to take adds the spurious solution $x = 1$. (This is an example of "extraneous solutions: multiplication", discussed at that page.) \begin{align*} (-x) \sqrt{4-(x-3)^2} &= (3-x) \sqrt{1-x^2} \\ (-x)^2 (4-(x-3)^2) &= (3-x)^2(1-x^2) \\ -5x^2 + 6x^3 - x^4 &= 9 - 6x -8x^2 +6x^3 -x^4 \\ 3x^2 + 6x -9 &= 0 \\ 3(x+3)(x-1) &= 0 \end{align*} So we found two solutions. Since in at least one step we multiplied through by an expresion that could be zero, we check all our solutions in the original equation. $$ \frac{3}{\sqrt{-8}} = \frac{-(-3)}{\sqrt{1-(-3)^2}} \overset{?}{=} \frac{3 - (-3)}{\sqrt{4-((-3)-3)^2}} = \frac{6}{\sqrt{-32}} $$ Since $\sqrt{-32} = \sqrt{4 \cdot -8} = 2 \sqrt{-8}$, we find that the two sides are equal and $-3$ is a solution to the original equation. $$ \frac{-1}{0} \unicode{x201C}\text{$=$"} \frac{-1}{\sqrt{1-1^2}} \overset{?}{=} \frac{3-1}{\sqrt{4-(1-3)^2}} \unicode{x201C}\text{$=$"} \frac{2}{0} $$ I put quotes around those equal signs because equality is a property of values and none of these expressions have a value because division by zero is undefined.
What happened? When $x = 1$, $\sqrt{1-x^2}$ of the two radicals we multiplied through by is zero, so we forced a spurious solution into our equation.
So, weirdly, the $x=-3$ is the correct solution. What could it possibly mean? Notice that at $x = -3$, the slope is complex valued. There is a sensible way to interpret a circle as a cmoplex object. So we could think of $x^2 + y^2 = 1$ where now $x$ and $y$ are complex variables. Unsurprisingly, it is hard to draw graphs of what is going on. But if you could, you would see that the complex slope of the complex tangent line to these two complex circles is the same for the points on the circles with $x$-coordinate equal to $-3 + 0\mathrm{i}$.
So what is the right solution technique?
Solving the first implicitly differentiated equation for its slope, we find $$ y_1' = \frac{-x_1}{y_1} $$ and so the point-slope form of a line tangent to the first circle is $y - y_1 = \frac{-x_1}{y_1}(x - x_1)$. Solving for slope-intercept form, $$ y = \frac{-x_1}{y_1}x + \left( \frac{x_1^2}{y_1} +y_1 \right) \text{.} $$ Repeating for the other circle, $$ y_2' = \frac{3-x_2}{y_2} \text{,} $$ $y - y_2 = \frac{3-x_2}{y_2}(x - x_2)$ and so $$ y = \frac{3-x_2}{y_2} x + \left( \frac{(x_2 - 3)x_2}{y_2} + y_2 \right) \text{.} $$
For these two lines to be the same line, they must have matching slopes and matching $y$-intercepts: \begin{align*} \frac{-x_1}{y_1} &= \frac{3-x_2}{y_2} \\ \frac{x_1^2}{y_1} +y_1 &= \frac{(x_2 - 3)x_2}{y_2} + y_2 \end{align*} Solve these with your two circle equations to find $x_1 = -1/3$, $y_1 = \frac{2\sqrt{2}}{3}$, $x_2 = 7/3$, and $y_2 = \frac{4\sqrt{2}}{3}$.
I guess you are lucky there is a unique solution. If the two circles did not have a point of tangency (at $(1,0)$), there could have been two doubly-tangent lines with positive slopes.