jmc

Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix. Completing the square on $x,$ we get $$y=-3(x+1{)}^2+3.$$To make the algebra a bit easier, we can find the focus of the parabola $y=-3x^2,$ shift the parabola left by 1 unit to get $y=-3(x+1{)}^2,$ and then shift it upward 3 units to find the focus of the parabola $y=-3(x+1{)}^2+3.$

Since the parabola $y=-3x^2$ is symmetric about the $y$-axis, the focus is at a point of the form $(0,f).$ Let $y=d$ be the equation of the directrix.



Let $(x,-3x^2)$ be a point on the parabola $y=-3x^2.$ Then $$P{F}^2=x^2+(-3x^2-f{)}^2$$and $P{Q}^2=(-3x^2-d{)}^2.$ Thus, $$x^2+(-3x^2-f{)}^2=(-3x^2-d{)}^2.$$Expanding, we get $$x^2+9x^4+6f{x}^2+f^2=9x^4+6d{x}^2+d^2.$$Matching coefficients, we get $$\begin{array}{rl}1+6f& =6d,\\ f^2& =d^2.\end{array}$$From the first equation, $d-f=\frac{1}{6}.$ Since $f^2=d^2,$ $f=d$ or $f=-d.$ We cannot have $f=d,$ so $f=-d.$ Then $-2f=\frac{1}{6},$ so $f=-\frac{1}{12}.$

Thus, the focus of $y=-3x^2$ is $\left(0,-\frac{1}{12}\right),$ and the focus of $y=-3(x+1{)}^2$ is $\left(-1,-\frac{1}{12}\right),$ so the focus of $y=-3(x-1{)}^2+3$ is $\boxed{\left(-1,\frac{35}{12}\right)}.$