你不能。在petsadopted中可以有两列：dogownerno和catownerno以及两个外键。但是table的设计似乎没有意义：不管是谁养的，宠物肯定不是狗就是猫（或者别的什么东西）？
这是另一种设计：

CREATE TABLE owner(
 ownerNo VARCHAR(8) CONSTRAINT ownerNo_pk2 PRIMARY KEY,
 ownerName VARCHAR(10)
);

CREATE TABLE petsAdopted(
 petNo VARCHAR(8) CONSTRAINT petNo_pk PRIMARY KEY,
 petType VARCHAR2(10) NOT NULL CONSTRAINT petTypeChk (CHECK petType in ('CAT','DOG'))
 ownerNo VARCHAR(8) CONSTRAINT ownerNo_fk REFERENCES owner(ownerNo)
);

这只是解决你的答案语法部分，你有很多语法错误，但这不会工作，请重新考虑你的设计。

CREATE TABLE dogOwner(
 ownerNo VARCHAR(8) CONSTRAINT ownerNo_pk1 PRIMARY KEY,
 ownerName VARCHAR(10)
);

CREATE TABLE catOwner(
 ownerNo VARCHAR(8),
 ownerName VARCHAR(10),
 CONSTRAINT ownerNo_pk2 PRIMARY KEY (ownerNo),
 CONSTRAINT ownerNo_fk1  FOREIGN KEY (ownerNo) REFERENCES dogOwner(ownerNo)
);

CREATE TABLE petsAdopted(
 petNo VARCHAR(8) ,
 ownerNo VARCHAR(8),
CONSTRAINT petNo_pk PRIMARY KEY (petNo),
CONSTRAINT ownerNo_fk_pet1 FOREIGN KEY (ownerNo) REFERENCES dogOwner(ownerNo),
CONSTRAINT ownerNo_fk_pet2  FOREIGN KEY (ownerNo) REFERENCES catOwner(ownerNo)
);