Try this: syms x y figure fimplicit(x^2 + y^2 -1, [-1 1]) axis('equal') .

Is this what you want: f = @(x,y) (x.^2 - y.^2).^2; % Your Function x = linspace(-10, 10, 150); % X-Range y = linspace(-10, 10, 150); % Y-Range [X,Y] = meshgrid(x,y); % Create Matrix Arguments figure(1) sh = surf(X, Y, f(X,Y)); set(sh, 'EdgeColor','none') grid on

ezplot('x^2/4+y^2/9=1'); axis equal; hold on ezplot('x^2+y^2=1')

fimplicit3(@(x,y,z) x.^2 +4*y.^2 +9*z.^2 - 1)

There are a few ways to go about this. One that is somewhat agnostic to what the equation is trying to represent (in this case, a circle) involves calculating the equation for the whole space, then plotting only an isoline of the target value. [X,Y] = meshgrid(-3:.1:3,-3:.1:3); % Generate domain. Z = X.^2 + Y.^2; % Find function value everywhere in the domain. contour(X,Y,Z,[4 4]) % Plot the isoline where the function value is 4. If you know more about your function and can turn it around into a function of only one variable (e.g., sine and cosine of t), that is preferable in most cases.

Try this: x = linspace(-2.0, 2.0, 50); y = linspace(-2.0, 2.0, 50); [xm, ym] = meshgrid(x, y); fxy = xm.^2 .* ym; surf(fxy); % Enlarge figure to full screen. set(gcf, 'units','normalized','outerposition',[0 0 1 1]); It gives you almost exactly what you want. You just have to rotate the viewpoint of the axis. I trust you can do this.