Step 1 :The given function is a quadratic function in two variables. The minimum or maximum of a quadratic function occurs at its vertex. The vertex of a quadratic function given in the form \(f(x, y) = ax^2 + by^2 + cx + dy + e\) is at the point \((-\frac{c}{2a}, -\frac{d}{2b})\). In this case, \(a = 1\), \(b = 3\), \(c = -4\), and \(d = 6\). So, the vertex is at \((\frac{4}{2}, -\frac{6}{2*3}) = (2, -1)\).
Step 2 :The temperature at this point is \(f(2, -1) = 2^2 + 3*(-1)^2 - 4*2 + 6*(-1)\).
Step 3 :The minimum temperature is -7 degrees Fahrenheit and it occurs at the point (2, -1).
Step 4 :Since the coefficients of \(x^2\) and \(y^2\) are positive, the function opens upwards and hence there is no maximum temperature.
Step 5 :The minimum temperature is \(\boxed{-7^{\circ} \mathrm{F}}\) at \((x, y)=\boxed{(2, -1)}\). There is no maximum temperature.