Construct the Jacobian matrix
Python, 63 bytes



lambda m,n:["df%d/du%%d "%i*n%tuple(range(n))for i in range(m)]


For m=3,n=2, outputs

['df0/du0 df0/du1 ', 'df1/du0 df1/du1 ', 'df2/du0 df2/du1 ']


The string formatting is 1 byte shorter than the more obvious

lambda m,n:[["df%d/du"%i+`j`for j in range(n)]for i in range(m)]