As one quick reference to how to do as Sylwester hints (using MIT/GNU Scheme instead of Racket. But the basic ideas are just same):

(define (sublist? lst1 lst2)
  (cond
    ;; recursion to find the possible starting location of lst1 inside lst2.
    ;; "starting" implies prefix?.
    ((prefix? lst1 lst2) #t)
    ((> (length lst1) (length lst2)) #f)
    (else (sublist? lst1 (cdr lst2))))
  )
(define (prefix? lst1 lst2)
  (cond
    ;; Here assume '() is always one prefix of any sublist.
    ((null? lst1) #t)
    ;; lst1 is non-null, but lst2 becomes null, so length of lst2 is less.
    ((null? lst2) #f)
    ;; base test
    ((equal? (car lst1) (car lst2)) (prefix? (cdr lst1) (cdr lst2)))
    ;; some elems in lst1 doesn't satisfy the base test
    (else #f)))

(sublist? '(a b d) '(a b c a b d e))
;Value: #t

(trace prefix?)
(prefix? '(a b) '(w a d f s))
; [Entering #[compound-procedure 13 prefix?]
;     Args: (a b)
;           (w a d f s)]
; [#f
;       <== #[compound-procedure 13 prefix?]
;     Args: (a b)
;           (w a d f s)]
;Value: #f