Try this (the letters of label are rotated but the aspect ratio remain constant so they can be read easily but are not look like "projected" on the xy plane)
The angles depend on figure/axes aspect ratio, so if you resize them you have to compute the angle again.
[X,Y,Z] = peaks(25);
fig = figure(1);
clf(fig);
t = tiledlayout(2,2);
for k = 1:prod(t.GridSize)
ax = nexttile(t);
surf(ax, X, Y, Z);
%axis(ax, 'equal')
view(ax, rand*360, 40); % random azimuth
set(ax,'unit','pixel');
axPosition = ax.Position;
dx = axPosition(3);
dy = axPosition(4);
[xx, xy] = ScreenProjection(ax, xlim(ax), [0 0], [0 0]);
thetax = atan(diff(xy)/diff(xx)*dy/dx);
xlabel('here is the x-label', 'Rotation', rad2deg(thetax))
[xx, xy] = ScreenProjection(ax, [0 0], ylim(ax), [0 0]);
thetay = atan(diff(xy)/diff(xx)*dy/dx);
ylabel('here is the y-label', 'Rotation', rad2deg(thetay))
zlabel('z')
end
% https://www.mathworks.com/matlabcentral/answers/430790-how-can-i-get-the-screen-coordinates-from-perspective-projection?s_tid=srchtitle
function [xcam, ycam] = ScreenProjection(ax, X, Y, Z)
dataRatio = get(ax, 'DataAspectRatio');
matrixRescale = diag(1./dataRatio);
CT = get(ax,'CameraTarget');
CP = get(ax,'CameraPosition');
CU = get(ax,'CameraUpVector');
cadeg = get(ax,'CameraViewAngle');
CT = CT(:);
CP = CP(:);
CU = CU(:);
ca = cadeg*pi/180;
%
c = CT-CP;
d = norm(c);
c = c / d;
u = CU - dot(c, CU)*c;
u = u/norm(u);
p = cross(c,u);
R=[c,p,u];
XYZ = [X(:),Y(:),Z(:)]';
CPU = R'*(matrixRescale*(XYZ-CP));
XYcam = CPU(2:3,:)./CPU(1,:);
xcam = XYcam(1,:);
ycam = XYcam(2,:);
end
