Plot the function we’re dealing with. Its lowest point is in the 1st quadrant of the xy plane; the ellipse we’re looking for extremes over extends to larger values in the other quadrants.

Plot3D[2x^2 + 3y^2 - 4x - 6y + 5, {x,0,3},{y,0,3}]

Mathematica’s Solve function solves an equation for the variable(s) you request, in this case L.

Solve[ (4/(4-2L))^2 + 2(6/(6-4L))^2 == 9, L ]

N[%2]

Calculate and save the x and y values corresponding to the real solutions.

x1 = 4/(4-2*0.6805988872968906)

y1 = 6 / (6 - 4*0.6805988872968906)

x2 = 4 / (4 - 2*2.795607979128593)

y2 = 6/(6 - 4*2.79560797912859)

Calculate the value of z at both (x, y) points found by the Lagrange multiplier. To save typing, define the formula for z as a user-defined Mathematica function, i.e., something you can call with different inputs at different times.

z[x_, y_] := 2x^2 + 3y^2 - 4x -6y + 5

z[1,0]

Finally, here are the minimum and maximum values of z on the ellipse:

z[x1,y1]

z[x2,y2]